Assignment 4 QA

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course Mth 163

004.

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Question: `q001. Note that this assignment has 4 questions

If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.

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Your solution:

F(3,7,-5) =(3, 13), (7, 53), (-5, 29)

A=1 b=0 c=4

X coordinate =0 y coordinate=4

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain

f(3) = 3^2 + 4 = 9 + 4 = 13.

Similarly we have

f(7) = 7^2 + 4 = 49 + 4 = 53 and

f(-5) = (-5)^2 + 9 = 25 + 4 = 29.

Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4.

The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.

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Question: `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.

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Your solution:

f(x) = x^2 + 4 also = f(a) = a^2 + 4.

When x is replaced by a

f(x+2) = (x + 2)^2 + 4 (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

This makes more sense on my paper than it does on here.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a:

f(a) = a^2 + 4.

Similarly to find f(x+2) we replace x with x + 2:

f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8.

To find f(x+h) we replace x with x + h to obtain

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4.

To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h):

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that

[ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.

You should have written these expressions out, and the following should probably be represented on your paper in form similar to that given here:

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Question: `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.

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Your solution:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

5 can be factored out leaving it to be 5( x2 - x1 )/( x2 - x1 )=5

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

Replacing x by the specified quantities we obtain the following:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

We can factor 5 out of the numerator to obtain

5 ( x2 - x1 ) / ( x2 - x1 ) = 5.

Compare what you have written down with the expressions below:

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Question: `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?

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Your solution:

5x+7=-3

-7=-7

5x=-10

X=-2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation

5x + 7 = -3.

We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.

Self-critique (if necessary):

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Self-critique rating:

Assignment 4 QA

#$&*

course Mth 163

004.

*********************************************

Question: `q001. Note that this assignment has 4 questions

If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F(3,7,-5) =(3, 13), (7, 53), (-5, 29)

A=1 b=0 c=4

X coordinate =0 y coordinate=4

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain

f(3) = 3^2 + 4 = 9 + 4 = 13.

Similarly we have

f(7) = 7^2 + 4 = 49 + 4 = 53 and

f(-5) = (-5)^2 + 9 = 25 + 4 = 29.

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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f(x) = x^2 + 4 also = f(a) = a^2 + 4.

When x is replaced by a

f(x+2) = (x + 2)^2 + 4 (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

This makes more sense on my paper than it does on here.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a:

f(a) = a^2 + 4.

Similarly to find f(x+2) we replace x with x + 2:

f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8.

To find f(x+h) we replace x with x + h to obtain

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4.

To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h):

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that

[ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.

You should have written these expressions out, and the following should probably be represented on your paper in form similar to that given here:

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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

5 can be factored out leaving it to be 5( x2 - x1 )/( x2 - x1 )=5

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

Replacing x by the specified quantities we obtain the following:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

We can factor 5 out of the numerator to obtain

5 ( x2 - x1 ) / ( x2 - x1 ) = 5.

Compare what you have written down with the expressions below:

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Question: `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

5x+7=-3

-7=-7

5x=-10

X=-2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation

5x + 7 = -3.

We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.

Self-critique (if necessary):

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Self-critique rating:

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Assignment 4 QA

Assignment 4 QA

@& This is a copy of a previous note. It appears that you're doing fine and understand this very well, but for the reason I explain below I can't review your work well enough to be sure.

@&
This is a copy of a previous note. It appears that you're doing fine and understand this very well, but for the reason I explain below I can't review your work well enough to be sure.