Expert: Scott A Wilson - 8/2/2010

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

  f(x) = -2x^2 + 4x + 16      f(x) = .67x^2 + 1.33x - 5.33
  f(x) = 2x^2 - 4x + 16      f(x) = -.67x^2 + 1.33x + 5.33

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

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

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

  f(x) = 4/5x2 + 2/5x - 48/5      f(x) = 2x2 + 4x - 48
  f(x) = 2/5x2 + 4/5x - 48/5      f(x) = 5x2 + 4x - 48

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

  f(x) = x2 - 6x      f(x) = -x2 - 6x
  f(x) = -x2 - 9x      f(x) = x2 - 6

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

  f(x) = -4/9x2 + 8/3      f(x) = 4/9x2 - 8/3x
  f(x) = 4/9x2 + 8/3x      f(x) = -4/9x2 - 8

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

  f(x) = x2 - 2x + 8      f(x) = x2 + 2x - 8
  f(x) = -x2 + 2x + 8      f(x) = -x2 - 2x + 8


On the first question I put a), b), c), and d), the same format follows throughout the rest of the questions. a)     c)
                   b)     d)

Answer
22. This says that we must have A(x+2)(x-4).  The average is at (-2+4)/2= 1.
We can put 1 in for x and -9A = 6, so divide both sides by -9.
Now that A is known, the equation can be multiplied out.

23. This says the equatin needs to be A(x+2)(x-3), and that when x is in the middle at
(-2+3)/2 = 1/2, that f(1/2) = -5.  That would be A(5/2)(-5/2) = -5, so 25A/4 = -5,
so A = -5*4/25 = -4/5.  That looks like d, since that is the only answer with fractions in it,
but check and make sure.

24. Again, f(x) = A(x+6)(x-4), f(-1) = -10, find A, multiply out ....

25. Max is in the middle of the interestion points; intersection points are at -6 and 0,
equation is f(x) = Ax(x+6); find A.

26. If x=6 is a zero and max is at x=3, then 6-3=3, so since 3 is 3 less than 6, the other intersection point is at 3 more than 6, so I'll say it is at B where B = 6+3.
The equations is then A(x-3)(x-B); B can be found, put x=3 in, find A to give that -4...

27. Determine the quadratic equation for the parabola:
range: {y:y ≤ 9};  x-intercepts: -2 and 4
This must be a parabola with a maximum at y=9 , and that is known to be at x=(-2+4)/2.
So take the equation y = A(x+2)(x-4), put in the x value just gotten, set that expression
equal to 9, and solve for A.