Expert: Scotto - 3/22/2010

Question
1. Determine the quadratic equation for the parabola:  
maximum value: 6;  x-intercepts: -2 and 4

a)f(x) = -2x^2 + 4x + 16  b)f(x) = .67x^2 + 1.33x - 5.33
c)f(x) = 2x^2 - 4x + 16  d)f(x) = -.67x^2 + 1.33x + 5.33


2. Determine the quadratic equation for the parabola:
 minimum value: -5;  x-intercepts: -2 and 3

a)f(x) = -4x^2 + 4x + 24  b)f(x) = 4x^2 - 4x - 24
c)f(x) = 5x^2 - 4x - 24  d)f(x) = 4/5x^2 - 4/5x - 24/5


3. Determine the quadratic equation for the parabola:  
  vertex: (-1,-10);  x-intercepts: -6 and 4

a)f(x) = 4/5x^2 + 2/5x - 48/5  b)f(x) = 2x^2 + 4x - 48
c)f(x) = 2/5x^2 + 4/5x - 48/5  d)f(x) = 5x^2 + 4x - 48



4. Determine the quadratic equation for the parabola:  
  maximum value: 9;  zeros of  f: -6 and 0

a)f(x) = x^2 - 6x  b)f(x) = -x^2 - 6x
c)f(x) = -x^2 - 9x  d)f(x) = x^2 - 6



5. Determine the quadratic equation for the parabola:  
  minimum value: -4 when x = 3;  zeros of  f: one is 6

a)f(x) = -4/9x^2 + 8/3  b)f(x) = 4/9x^2 - 8/3x
c)f(x) = 4/9x^2 + 8/3x  d)f(x) = -4/9x^2 - 8



6. Determine the quadratic equation for the parabola:  
  range: {y:y ≤ 9};  x-intercepts: -2 and 4

a)f(x) = x^2 - 2x + 8  b)f(x) = x^2 + 2x - 8
c)f(x) = -x^2 + 2x + 8  d)f(x) = -x^2 - 2x + 8

Answer
1. Intercepts at -2 and 4 imply that it is y=C(x+2)(x-4) => y = C(x²-2x-8) where C is a constant.

The maximum is when the x value is halfway into between -2 and 4, so the max is at x = 1.
Putting in x=1 gives -9C, and that is 6, so C = -2/3.
Take C and put it in y = C(x²-2x-8) to get the correct answer.


2. The conditions imply the equation is y = C(x+2)(x-3).
The maximum value would be at x = (-2+3)/2 = 1/2.
Given that C(x+2)(x-3) = 5 at x = 1/2, find C.


3. The equation is y = C(x+6)(x-4), since the intercepts are at -6 and 4.
The vertex determines C by going -10 = C(-1+5)(-1-4) and solving for C.


4. The equation would be y = Cx(x+6).  The max is halfway between the roots, at x = -3.
Putting in x=-3 and getting 9 determine C.


5. Since one of the roots is at x=6 and the vertex is at x=3, and the vertex is half way between the two roots, this means the other is 6-3 = 3 points below 3, which is at 0.
Thus, the equation is y = Cx(x-6).  Take 3 and put it in, giving -4.
That is, -4 = 3C(x-6), and solve for C.


6. This means the max is at y=9 and the equation is y = C(x+2)(x-4)
The vertex is halfway between -2 and 4, so it is at x = 1.
So put in x=1 into the equation and note that y=9 at this point.