Assignment 30 45

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course mth 158

8/2/13 6:39pmI am working on getting everything done and in before the end of the class period!!

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

030. * 30

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Question: * 4.3.36 / 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.

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Your solution:

f(x)= x^2-2x-3

a = 1 b = -2 c = -3

h = -b/2a k = f(-b/2a)

h = -b/2a

h = - (-2)/2

h = 1

k = f(1)

f (1) = (1)^2 - 2 (1) -2

= 1-2-3

= -4

so vertex (h, k) = (1, -4)

Axis line x = 1

a = 1 and 1 > 0 so opens up

Intercepts

f (0) = 0-0-3

f (0) = -3

so y-int = (0, -3)

0 = x^2 - 2x - 3

= (x + 1) (x-3)

x = -1 x = 3

so x-ints = (-1, 0) and (3, 0)

domain is all real numbers

range confuses me a bit, however, and I am not sure that I remember how to do it for this

confidence rating #$&*: 2

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Given Solution:

* * The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3.

The graph of this quadratic function will open upwards, since a > 0.

The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4).

The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get

(x - 3) ( x + 1) = 0 so that

x - 3 = 0 OR x + 1 = 0, giving us

x = 3 OR x = -1.

So the x intercepts are (-1, 0) and (3, 0).

The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).

The function can be evaluated for any real number x, so its domain is the set of all real numbers.

The graph is a parabola opening upward, with vertex (1, -4). So the lowest possible y value is -4, and all values greater than -4 will occur. So the range is y > -4.

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Self-critique (if necessary):

The explanation helped me to remember the proper way to find the range. It is much easier to look in the graph and see the range than to try to determine it from the problem itself, which I have some trouble with at times.

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Self-critique Rating: 3

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Question: * 4.3.51 / 4.3.57. graph of parabola vertex (1, -3), point (3, 5)

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Your solution:

h = 1 k = -3 x = 3 y / f(x) = 5

f (x) = a (x-h)^2 + k

5 = a (3 - 1)^2 - 3

5 = 16a -3

8 = 16a

2 = a

f (x) = 2 (x - 1)^2 - 3

confidence rating #$&*: 3

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Given Solution:

* * The graph is a parabola with vertex (1, -3), so it has form

(y - k) = a ( x - h)^2, with h = 1 and k = -3.

Thus we have

y - (-3) = a (x - 1)^2,

and the only unknown is a. To find a we substitute the coordinates of (3, 5) for x and y, obtaining the equation

(5 - (-3)) = a ( 3 - 1)^2, or{}(5 + 3) = a * 2^2, or{}8 = a * 4,

with the obvious solution

a = 2.

Thus the equation of the parabola is

y + 3 = 2 ( x - 1)^2.

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Self-critique (if necessary): OK, what I wrote should be equivalent

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Self-critique Rating: OK

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Question: * Extra Problem (4.1.67 7th edition) (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?

Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3).

If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15.

If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30.

If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30.

If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.

Does the value of a affect the location of the vertex?

In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.

The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following:

For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12.

For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24.

For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24.

For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60.

So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).

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Self-critique (if necessary):

I do not really understand how to do this problem; I am not sure how to get a quadratic function just from the a value and the x-intercepts. I kind of see in the explanation how the given information was plugged in, but doesn't that leave out parts of the quadratic function formula f(x) = a (x-h)^2 + k? The more I look at this the more it confuses me.

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Self-critique Rating: 1

@&
As you say the general form of the quadratic function is

f(x) = a (x-h)^2 + k.

There are other forms, each with its own purpose, and one form can be connected to another by basic algebra.

The present form is convenient for graphing the function, since it tells us the vertex (h, k) and the steepness of the parabola, which is determined by a.

If we expand this first form we get

f(x) = a ( x^2 - 2 x h + h^2) + k
= a x^2 - 2 a x h + a h^2 + k.

Note the coefficient of x^2. That coefficient is a.

The equation of a quadratic function can also be written as

f(x) = a x^2 + b x + c.

Again the coefficient of x is a.

The thing to understand about this form is that it can often be factored, and if it can the factors tell us where the value of the function will be zero.

Now, if a quadratic function has zeros at x = z1 and x = z2, the quadratic function must be a multiple of

(x - z1) * (x - z2).

If the x^2 term is to have coefficient a, the specific multiple must be

a ( x - z1) * (x - z2).

The zeros correspond to the x intercepts, so in the case of these functions the zeros are -5 and 3, so the form of the function must be

a ( x - (-5) ) * (x - 3), or
a ( x + 5) ( x - 3).

Thus:

If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15.
If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30.
If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30.
If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
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Assignment 30 45

&#Your work looks good. See my notes. Let me know if you have any questions. &#