Assignment 003

course Mth 163

???????????h??assignment #003003. `query 3

Precalculus I

09-16-2008

......!!!!!!!!...................................

22:50:56

query graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.

......!!!!!!!!...................................

RESPONSE -->

If the graph y = x^2 was strecthed

Y = x^2 + 3 the y axis would rise by 3 measures

Y = x^2 + 2 the y axis would rise by 2 measures

Y = x^2 + .5 the y axis would rise by .5 measures

Y = x^2 + -3 the y axis would decrease by -3 measures

confidence assessment: 1

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

......!!!!!!!!...................................

22:56:47

** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up.

INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). **

......!!!!!!!!...................................

RESPONSE -->

I'm not sure if this is what I have, but I did forget to explain how the paribola would expand or contract according to the direction of stretch.

self critique assessment: 2

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

......!!!!!!!!...................................

23:41:34

query prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1

......!!!!!!!!...................................

RESPONSE -->

I think I have it

y = x^2 + 2x + 1

-1 = x^2 + 2x

This would result from the equation

0 = x^2 + 2x + 1,

but there is no justification for the substitution y = 0.

If we did wish to solve this equation, which would tell us the values of x at which the graph crosses the x-axis, we would use the quadratic formula. The equation cannot be solved by the type of rearrangement you attempt below.

You can't get from

= -1/1 = 2^2/(4 * 1^2)
-1 = x^2 + (2/1)

You can't get from

= -1/1 = 2^2/(4 * 1^2) to
-1 = x^2 + (2/1)

by adding, subtracting, multiplying or dividing both sides by the same quantity.

x = (-2) +- sqrt ( 2^2 -4 ) / 2 = -2

The quadratic formula will work to find the x intercepts. For this equation the quadratic formula would give you

x = ((-2) +- sqrt ( 2^2 -4 )) / 2 = (-2 +- sqrt(0) ) = -2 / 2 = -1.

It turns out that both x intercepts are at the same point, which means that in this case the vertex is on the x-axis.

x = -2

So in the first equation y = 1

And x = -2

confidence assessment: 2

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

......!!!!!!!!...................................

23:42:46

** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a).

For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0).

The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1).

For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4).

The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, 1.25) and (-.5, 1.25). **

......!!!!!!!!...................................

RESPONSE -->

Isn't that what I have?

Not quite. See my notes above.

self critique assessment: 2

you don't have your vertex at x = -1.

If x = -2, you are correct that for this function y = 1, but that will not be the vertex.

The vertex is located at x = - b / (2 a).

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

......!!!!!!!!...................................

23:45:00

how did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?

......!!!!!!!!...................................

RESPONSE -->

According to my graphs the vertex moved from right to left.

confidence assessment: 1

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

......!!!!!!!!...................................

23:45:22

** The vertices move downward and to the left, but not along a straight line.

In fact the vertices lie along a different parabola of their own. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 2

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

......!!!!!!!!...................................

23:46:32

How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?

......!!!!!!!!...................................

RESPONSE -->

By using the quadratic formula is the only way I can even come close.

confidence assessment: 1

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

......!!!!!!!!...................................

23:47:31

** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symetry around the vertex which defines the 'shape' and direction and allows you to extrapolate.

INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. **

......!!!!!!!!...................................

RESPONSE -->

I am still haveing trouble understanding the formula.

self critique assessment: 1

hopefully my notes will be helpful. You should be familiar with the quadratic formula from prerequisite courses, but in any case you appear to be pretty close to using it correctly.

The vertex is located along the line x = - b / (2 a), which lies halfway between the zeros of the function, has given by the quadratic formula.

The other two basic points lie 1 unit to the right and left of the vertex, and a units vertically from the vertex.

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

......!!!!!!!!...................................

23:48:16

query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?

......!!!!!!!!...................................

RESPONSE -->

c and a

confidence assessment: 2

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

......!!!!!!!!...................................

23:51:06

** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present

INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros.

The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). **

......!!!!!!!!...................................

RESPONSE -->

If c = 0 then wouldn't one of the points be 0

and if the last part of the formula if a = 0 then wouldn't the x=0?

self critique assessment: 2

If c = 0 then (0, 0) would lie on the graph.

If a = 0 then the function is of the form y = b x + c, which is a straight line with slope b and y intercept c. Unless c = 0, this function will not be zero when x = 0.

However the question asks about general values of a, b and c, and the thing that determines the nature of the zeros is the discriminant, as given in the above solution.

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

......!!!!!!!!...................................

23:57:24

query #4. Questions about vertex between zeros and the shape of the curve connecting vertices:

What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?

......!!!!!!!!...................................

RESPONSE -->

If the y axis is negitive the there will be 2 0's on the x axis

confidence assessment: 2

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

......!!!!!!!!...................................

23:57:45

** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 2

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

......!!!!!!!!...................................

23:58:42

What was the shape of the curve connecting the vertices?

......!!!!!!!!...................................

RESPONSE -->

Always accending at a increasing rate

confidence assessment: 1

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

......!!!!!!!!...................................

23:59:32

** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea.

Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 1

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

......!!!!!!!!...................................

00:00:49

Query Add comments on any surprises or insights you experienced as a result of this assignment.

......!!!!!!!!...................................

RESPONSE -->

I have a little better understanding of quadratic formula but have still not mastered it.

self critique assessment: 3

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

See my notes, which are fairly extensive, and let me know if you have questions. You are welcome to insert modified solutions, questions, or comments into this document, marked by $$$$, and submit a copy.

Assignment 003

You can't get from = -1/1 = 2^2/(4 * 1^2) -1 = x^2 + (2/1)

You can't get from = -1/1 = 2^2/(4 * 1^2) to -1 = x^2 + (2/1) by adding, subtracting, multiplying or dividing both sides by the same quantity.

The quadratic formula will work to find the x intercepts. For this equation the quadratic formula would give you x = ((-2) +- sqrt ( 2^2 -4 )) / 2 = (-2 +- sqrt(0) ) = -2 / 2 = -1. It turns out that both x intercepts are at the same point, which means that in this case the vertex is on the x-axis.

Not quite. See my notes above.

you don't have your vertex at x = -1. If x = -2, you are correct that for this function y = 1, but that will not be the vertex. The vertex is located at x = - b / (2 a).

See my notes, which are fairly extensive, and let me know if you have questions. You are welcome to insert modified solutions, questions, or comments into this document, marked by $$$$, and submit a copy.

You can't get from

= -1/1 = 2^2/(4 * 1^2)
-1 = x^2 + (2/1)

You can't get from

= -1/1 = 2^2/(4 * 1^2) to
-1 = x^2 + (2/1)

by adding, subtracting, multiplying or dividing both sides by the same quantity.

you don't have your vertex at x = -1.

If x = -2, you are correct that for this function y = 1, but that will not be the vertex.

The vertex is located at x = - b / (2 a).