Question

Which of the following is the vertex of the quadratic equation y = –4(x + 6)^2 + 2?

(6, 2)

(–6, 2)

(6, –2)

(–6, –2)

Answer 1

its (-6,2)

Answer 2

For y = a(x -h)^2 + k, the vertex is (h,k).

Answer 3

\[y'=-8(x+6)\]The vertex is the critical point of the parabola.\[0=-8(x+6)\\0=x+6\\x=-6\]Now, just determine the y.\[y=-4(-6+6)^2+2=2\]So the vertex is...\[(-6,2)\]

Answer 4

Or, the easy way to do it... we know the vertical parabola is in vertex form i.e.\[y-k = a(x-h)^2\]... then the vertex is:\[(h,k)\]In our case, we find the variables are as follows.\[y=4(x+6)^2+2\\y-2=4(x- -6)^2\]\[a=4\\h=-6\\k=2\]So, the vertex is then...\[(h,k)=(-6,2)\]

Answer 5

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