Assignment 28

course Mth 158

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028. Query 28

College Algebra

08-08-2008

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06:17:30

4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.

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RESPONSE -->

The graph opens upward, vertex is (3,-2), axis of symmetry is 3, x-intercepts (1,0) (5,0)

Domain x>= -2

confidence assessment: 2

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06:18:36

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).

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RESPONSE -->

I tried to use just the graph which was wrong..I think I understand.

self critique assessment: 3

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06:29:04

4.1.67 (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?

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RESPONSE -->

Book problem was intercepts -3 and 1

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

f(x)=2(x+3)(x-1)=2x^2-4x+6

f(x)=-2(x+3)(x-1)=-2x^2-4x+6

f(x)=5(x+3) (x-1)=5x^2+10x-15

confidence assessment: 3

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06:29:20

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.

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RESPONSE -->

It looks like the same form.

self critique assessment: 3

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06:30:39

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

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RESPONSE -->

no the vertex will remain in the same place halfway between the x-intercepts

confidence assessment: 1

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06:32:50

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.

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RESPONSE -->

ok

self critique assessment: 3

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06:34:12

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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RESPONSE -->

ok

self critique assessment: 3

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06:44:39

4.1.78 (was 4.1.60). A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

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RESPONSE -->

two sides of the rectangle would =x and the other two sides would equal 2000-x

x(2000-x)/2=2000x-x^2/a=-x^2/2+1000x

vertex at x=-1000/2(-1/2)

If x=1000

-x^2/2+1000x=-1000^2/2+1000(1000)=500,000

So 500,000 would be the most he could fence without fencing the side closest to the highway.

confidence assessment: 2

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06:44:54

If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters.

The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x.

The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000.

At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters.

Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **

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RESPONSE -->

Looks correct.

self critique assessment: 3

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06:53:31

4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle.

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RESPONSE -->

x(10-x)=10x-x^2 or -x^2+10x

vertex =-10/2 (-1)=5

-(5^2)+10 (5)=-25+50=25

25 is the maximum A of the rectangle.

confidence assessment: 2

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06:53:40

The dimensions of the rectangle are x and y = 10 - x. So the area is

area = x ( 10 - x) = -x^2 + 10 x.

The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5.

Since the parabola opens downward this value of x results in a maximum area, which is

-x^2 + 10 x = -5^2 + 10 * 5 = 25. **

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RESPONSE -->

ok

self critique assessment: 3

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