Assignment 2 QA

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course Mth 163

Question `q001: Sketch a set of coordinate axes, with the x axis horizontal and directed to the right, the y axis vertical and directed upwards.Sketch the point P = (-3, -1) on a set of coordinate axes.

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Shifting -1 = (-4,1)

Shifting 3 = (-3,2)

Shifting 4 = (-3,-4)

Final Point (-4,-1)

confidence rating #$&*:

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2

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (-3 + (-1), -1) = (-4, -1).

Shifting the point 3 units in the vertical direction we end up at the point (-3, -1 + ( 3)) = (-3, 2).

The point (-3, -1) is -1 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will become 4 * -1 = -4.

• The x coordinate will not change.

• So the coordinates of the new point will be (-3, -4).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-3 + (-1), -4) = (-4, -4).

If you shift this new point 3 units in the vertical direction, it will end up at (-4, -4 + 3) = (-4, -1).

NOTE: We can express this sequence of transformations in a single step as

(-3 + (-1), 4 * -1 + 3) = (-4, -1).

Question `q002: Starting with the point P = (0, 0):

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Shifting -1 = (-1,0)

Shifting 3 = (0,3)

Shifting 4 = (0,0)

Final Point = (-1,3)

confidence rating #$&*:

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2

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0).

Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3).

The point (0, 0) is 0 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0.

• The x coordinate will not change.

• So the coordinates of the new point will be (0, 0).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0).

If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3)

Question `q003: Plot the points (0, 0), (-1, 1) and (1, 1) on a set of coordinate axes.

Now plot the points you get if you move each of these points 4 times further from the x axis, and put a small circle around each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three circled points -1 units in the x direction. Put a small 'x' through each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three new points (the ones with the x's) 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

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Multiply by 4 (-1,4) (0,0) (1,4)

-1 direction (-2,4) (-1,0) (0,4)

3 (-2,7) (-1,3) (0,7)

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Given Solution:

2

Moving each point 4 times further from the x axis:

The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is.

The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4).

At this stage our three points are

• (-1, 4)

• (0, 0)

• (1, 4)

Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points

(-1 + (-1), 4) = (-2, 4),

(0 + -1, 0) = (-1, 0) and

((1 + (-1), 4) = ( 0, 4), so our points are now

• (-2, 4)

• (-1, 0)

• ( 0, 4)

Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points

(-2, 4 + 3) = (-2, 7)

(-1, 0 + 3) = (-1, 3) and

( 0, 4 + 3) = ( 0, 7)

Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1).

Then sketch the parabola corresponding to your three circled basic points.

Then sketch the parabola corresponding to three basic points you indicated with 'x's'.

Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'.

Describe how each parabola is related to the one before it.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

confidence rating #$&*:

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Given Solution: Your 'circled-points' parabola will be narrower than the original parabola through (-1, 1), (0, 0) and (1, 1). In fact, each point on the 'circled-points' parabola will lie 4 times further from the x axis than the point on the original parabola.

Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction.

Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction.

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Question: `q005.

Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables:

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C is the easiest.

confidence rating #$&*:

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3

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Given Solution:

The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results.

Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is

• 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b.

The right-hand side will be the difference 90 - 128 = -38, so the second equation will become

• new' 2d equation: 58 a + 2 b = -38.

The 'new' third equation by a similar calculation will be

• 'new' third equation: 198 a + 7 b = -128.

You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second).

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Question: `q006. Solve the two equations

58 a + 2 b = -38

198 a + 7 b = -128

which can be obtained from the system in the preceding problem, by eliminating the easiest variable.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I would say B therefor a=1

confidence rating #$&*:

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2

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Given Solution:

Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a.

To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications:

-7 * ( 58 a + 2 b) = -7 * -38

2 * ( 198 a + 7 b ) = 2 * (-128)

Doing the arithmetic we obtain

-406 a - 14 b = 266

396 a + 14 b = -256.

Adding the two equations we obtain

-10 a = 10,

so we have

a = -1.

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Question: `q007. Having obtained a = -1, use either of the equations

58 a + 2 b = -38

198 a + 7 b = -128

to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

58 * -1 + 2b=-38

-58+2b=-38

2b=-38+58

2b=20

B=10

confidence rating #$&*:

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3

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Given Solution:

You might have completed this step in your solution to the preceding problem.

Substituting a = -1 into the first equation we have

58 * -1 + 2 b = -38, so

2 b = 20 and

b = 10.

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Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2*-1+3*10+c=128

28+c=128

-28=-28

C=100

confidence rating #$&*:

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3

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Given Solution:

Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100.

Substituting these values into the second equation, in order to check our solution, we obtain

60 * -1 + 5 * 10 + 100 = 90, or

-60 + 50 + 100 = 90, or

90 = 90.

We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution.

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Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8).

We are going to use simultaneous equations to obtain the equation of that parabola.

• A graph has a parabolic shape if its the equation of the graph is quadratic.

• The equation of a graph is quadratic if it has the form y = a x^2 + b x + c.

• y = a x^2 + b x + c is said to be a quadratic function of x.

To find the precise quadratic function that fits our points, we need only determine the values of a, b and c.

• As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c.

The first step is to obtain an equation using the first known point.

• What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A+b+c=-2

confidence rating #$&*:

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2

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Given Solution:

We substitute y = -2 and x = 1 to obtain the equation

-2 = a * 1^2 + b * 1 + c, or

a + b + c = -2.

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Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

9a+3b+c=5 49a+7b+c=8

confidence rating #$&*:

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3

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Given Solution:

Using the second point we substitute y = 5 and x = 3 to obtain the equation

5 = a * 3^2 + b * 3 + c, or

9 a + 3 b + c = 5.

Using the third point we substitute y = 8 and x = 7 to obtain the equation

8 = a * 7^2 + b * 7 + c, or

49 a + 7 b + c = 8.

Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this.

Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c.

Solve the system to find the values of a, b and c.

• What is the solution of this system?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a = - 0.45833 b = 5.33333 c = - 6.875

confidence rating #$&*:

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Given Solution:

The system consists of the three equations obtained in the last problem:

a + b + c = -2

9 a + 3 b + c = 5

49 a + 7 b + c = 8.

This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers.

The solution of this system, in decimal form, is approximately

a = - 0.45833,

b = 5.33333 and

c = - 6.875.

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Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function.

• What is your function?

• What y values do you get when you substitute x = 1, 3, 5 and 7 into this function?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=-0.45833x^2+5.33333x-6.875

Y= -2,5,8.33333,8

Which equals (1,-2) (3,5) 5,8.3333) and (7,8)

confidence rating #$&*:

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Given Solution:

Substituting the values of a, b and c into the given form we obtain the equation

y = - 0.45833 x^2 + 5.33333 x - 6.875.

• When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2.

• When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5.

• When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333.

• When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8.

Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to

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Self-critique (if necessary):

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Self-critique rating:

Assignment 2 QA

#$&*

course Mth 163

Question `q001: Sketch a set of coordinate axes, with the x axis horizontal and directed to the right, the y axis vertical and directed upwards.Sketch the point P = (-3, -1) on a set of coordinate axes.

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Shifting -1 = (-4,1)

Shifting 3 = (-3,2)

Shifting 4 = (-3,-4)

Final Point (-4,-1)

confidence rating #$&*:

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2

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (-3 + (-1), -1) = (-4, -1).

Shifting the point 3 units in the vertical direction we end up at the point (-3, -1 + ( 3)) = (-3, 2).

The point (-3, -1) is -1 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will become 4 * -1 = -4.

• The x coordinate will not change.

• So the coordinates of the new point will be (-3, -4).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-3 + (-1), -4) = (-4, -4).

If you shift this new point 3 units in the vertical direction, it will end up at (-4, -4 + 3) = (-4, -1).

NOTE: We can express this sequence of transformations in a single step as

(-3 + (-1), 4 * -1 + 3) = (-4, -1).

Question `q002: Starting with the point P = (0, 0):

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Shifting -1 = (-1,0)

Shifting 3 = (0,3)

Shifting 4 = (0,0)

Final Point = (-1,3)

confidence rating #$&*:

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2

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0).

Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3).

The point (0, 0) is 0 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0.

• The x coordinate will not change.

• So the coordinates of the new point will be (0, 0).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0).

If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3)

Question `q003: Plot the points (0, 0), (-1, 1) and (1, 1) on a set of coordinate axes.

Now plot the points you get if you move each of these points 4 times further from the x axis, and put a small circle around each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three circled points -1 units in the x direction. Put a small 'x' through each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three new points (the ones with the x's) 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

confidence rating #$&*:

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Multiply by 4 (-1,4) (0,0) (1,4)

-1 direction (-2,4) (-1,0) (0,4)

3 (-2,7) (-1,3) (0,7)

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Given Solution:

2

Moving each point 4 times further from the x axis:

The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is.

The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4).

At this stage our three points are

• (-1, 4)

• (0, 0)

• (1, 4)

Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points

(-1 + (-1), 4) = (-2, 4),

(0 + -1, 0) = (-1, 0) and

((1 + (-1), 4) = ( 0, 4), so our points are now

• (-2, 4)

• (-1, 0)

• ( 0, 4)

Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points

(-2, 4 + 3) = (-2, 7)

(-1, 0 + 3) = (-1, 3) and

( 0, 4 + 3) = ( 0, 7)

Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1).

Then sketch the parabola corresponding to your three circled basic points.

Then sketch the parabola corresponding to three basic points you indicated with 'x's'.

Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'.

Describe how each parabola is related to the one before it.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

confidence rating #$&*:

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Given Solution: Your 'circled-points' parabola will be narrower than the original parabola through (-1, 1), (0, 0) and (1, 1). In fact, each point on the 'circled-points' parabola will lie 4 times further from the x axis than the point on the original parabola.

Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction.

Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction.

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Question: `q005.

Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables:

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C is the easiest.

confidence rating #$&*:

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3

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Given Solution:

The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results.

Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is

• 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b.

The right-hand side will be the difference 90 - 128 = -38, so the second equation will become

• new' 2d equation: 58 a + 2 b = -38.

The 'new' third equation by a similar calculation will be

• 'new' third equation: 198 a + 7 b = -128.

You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second).

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Question: `q006. Solve the two equations

58 a + 2 b = -38

198 a + 7 b = -128

which can be obtained from the system in the preceding problem, by eliminating the easiest variable.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I would say B therefor a=1

confidence rating #$&*:

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2

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Given Solution:

Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a.

To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications:

-7 * ( 58 a + 2 b) = -7 * -38

2 * ( 198 a + 7 b ) = 2 * (-128)

Doing the arithmetic we obtain

-406 a - 14 b = 266

396 a + 14 b = -256.

Adding the two equations we obtain

-10 a = 10,

so we have

a = -1.

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Question: `q007. Having obtained a = -1, use either of the equations

58 a + 2 b = -38

198 a + 7 b = -128

to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

58 * -1 + 2b=-38

-58+2b=-38

2b=-38+58

2b=20

B=10

confidence rating #$&*:

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3

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Given Solution:

You might have completed this step in your solution to the preceding problem.

Substituting a = -1 into the first equation we have

58 * -1 + 2 b = -38, so

2 b = 20 and

b = 10.

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Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2*-1+3*10+c=128

28+c=128

-28=-28

C=100

confidence rating #$&*:

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3

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Given Solution:

Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100.

Substituting these values into the second equation, in order to check our solution, we obtain

60 * -1 + 5 * 10 + 100 = 90, or

-60 + 50 + 100 = 90, or

90 = 90.

We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution.

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Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8).

We are going to use simultaneous equations to obtain the equation of that parabola.

• A graph has a parabolic shape if its the equation of the graph is quadratic.

• The equation of a graph is quadratic if it has the form y = a x^2 + b x + c.

• y = a x^2 + b x + c is said to be a quadratic function of x.

To find the precise quadratic function that fits our points, we need only determine the values of a, b and c.

• As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c.

The first step is to obtain an equation using the first known point.

• What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?

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Your solution:

A+b+c=-2

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2

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Given Solution:

We substitute y = -2 and x = 1 to obtain the equation

-2 = a * 1^2 + b * 1 + c, or

a + b + c = -2.

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Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation)

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Your solution:

9a+3b+c=5 49a+7b+c=8

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3

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Given Solution:

Using the second point we substitute y = 5 and x = 3 to obtain the equation

5 = a * 3^2 + b * 3 + c, or

9 a + 3 b + c = 5.

Using the third point we substitute y = 8 and x = 7 to obtain the equation

8 = a * 7^2 + b * 7 + c, or

49 a + 7 b + c = 8.

Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this.

Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c.

Solve the system to find the values of a, b and c.

• What is the solution of this system?

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Your solution:

a = - 0.45833 b = 5.33333 c = - 6.875

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1

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Given Solution:

The system consists of the three equations obtained in the last problem:

a + b + c = -2

9 a + 3 b + c = 5

49 a + 7 b + c = 8.

This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers.

The solution of this system, in decimal form, is approximately

a = - 0.45833,

b = 5.33333 and

c = - 6.875.

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Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function.

• What is your function?

• What y values do you get when you substitute x = 1, 3, 5 and 7 into this function?

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Your solution:

Y=-0.45833x^2+5.33333x-6.875

Y= -2,5,8.33333,8

Which equals (1,-2) (3,5) 5,8.3333) and (7,8)

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2

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Given Solution:

Substituting the values of a, b and c into the given form we obtain the equation

y = - 0.45833 x^2 + 5.33333 x - 6.875.

• When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2.

• When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5.

• When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333.

• When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8.

Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to

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&#Very good responses. Let me know if you have questions. &#