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)

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)

anonymous · Answer

its (-6,2)

hartnn · Answer

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

anonymous · Answer

$$y'=-8(x+6)$$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...$$(-6,2)$$

anonymous · Answer

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...$$(h,k)=(-6,2)$$

hartnn · Answer

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