Respuesta :
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.
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:
D.The vertex form is f(x) = −(x +3)^2 + 4. The standard form is f(x) = −x2−6x−5.
Step-by-step explanation:
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