Question

Find the equation, (f(x) = a(x-h)2+ k), for a parabola containing point (-1,0) and having (-3, 4) as a vertex. What is the standard form of the equation?

Answer 1

Let's begin with f(x) = a(x-h)^2+ k. Note that we must use "^" to indicate exponentiation. Write (x-h)^2, not (x-h)2.

If (-3,4) is the vertex, then the above equation becomes f(x) = a(x-[-3])^2 + 4, or

f(x) = a(x+3)^2 + 4. We are told that the graph passes through (-1,0), so must now substitute those coordinates into the above equation:

f(-1) = a([-1]+3)^2 + 4 = 0 (0 is the value of f when x is -1)

Then we have a(2)^2 + 4 =0, or 4a + 4 = 0. Thus, a = -1.

The equation of this parabola is now f(x) = -(x+3)^2 + 4.

Write it in "standard form:" f(x) = -(x^2 + 6x + 9) + 4, or
f(x) = -x^2 - 6x - 9 + 4, or
answer => f(x) = -x^2 - 6x - 5 = ax^2 + bx + c

Thus, a=-1, b=-6 and c= -5.

Answer 2

D.The vertex form is f(x) = −(x +3)^2 + 4. The standard form is f(x) = −x2−6x−5.

Explanation:

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