Assignment 30

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course Mth 158

4/12/13 Around 1:30PM

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

 1 > 0, parabola opens up

 - b/(2a) = - (-2)/(2*1) = 1

 f(- b/(2a)) = x^2 - 2x - 3

 f(1) = (1)^2 - 2(1) - 3

 = 1 - 2 - 3

 = -4

 (1 , -4) = vertex

axis of symmetry = (-1 , 0)

 b^2 - 4ac = (-2)^2 - 4(1)(-3) = 16 > 0, two x-int, crosses x axis in two places

 (0)^2 - 2(0) - 3 = -3, y-int = (0 , -3)

 x^2 - 2x - 3 = 0

 (x - 3)(x + 1) = 0, x-int = (3 , 0), (-1 , 0)

 Domain = All real numbers

 Range = [-4 , infinity)

 Increasing (-4 , infinity)

 Decreasing (infinity , -4)

 

 

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): OK

 

 

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Self-critique Rating:

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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:

 5 = a(3 - 1)^2 - 3

 5 = 4a - 3

 8 = 4a

 2 = a

 f(x) = 2(x - 1)^2 - 3

 = 2(x^2 - 2x + 1) - 3

 = 2x^2 - 4x - 1

 

 

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):

 I used the equation: f(x) = a(x - h)^2 + k, to get ""a"" because that's how it was taught in my text book.

 

 

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Self-critique Rating:

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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 could not complete the problem because I didn't understand where to go from ""What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?"" and since I have the ninth edition of the textbook, I couldn't look up the problem. I believe I see how it works though.

 

 

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Self-critique Rating: 2

 

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