Find the vertex for the parabola whose equation is given.

y = -4x^2 + 8x + 8


a) (-1, -4)
b) (-2, -24)
c) (1, 12)
d) (2, 0)

  C)  (1, 12)

Step-by-step explanation:

You can rewrite the equation in vertex form:

  y = -4(x^2 -2x) +8 . . . . . factor the leading coefficient from the first two terms

  y = -4(x^2 -2x +1) +8 +4 . . . . .  add the square of half the x-coefficient inside parentheses (-4)((-2/2)^2) and its opposite outside parentheses (-(-4)).

Now, the equation can be written in vertex form.

  y = a(x -h)^2 +k . . . . . . . . . has vertex (h, k)

  y = -4(x -1)^2 +12 . . . . . . . . has vertex (1, 12)

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Comment on the problem

The typewritten form of the equation seems to have a missing leading minus sign. Its vertex is (-1, 4), so "none of the above."

step-by-step explanation:

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