Question

Can someone explain to me how to find the vertex form of this equation? y= -x^2 + 12x-4

Answer 1

ANSWER:  

The vertex of given equation  $y=-x^{2}+12 x-4$ is (6, 32).

SOLUTION:

Given, quadratic equation is  $y=-x^{2}+12 x-4$

Above equation is a parabola and we need to find the vertex of that parabola.  We know that, general form of the parabola is  $y=a(x-h)^{2}+k$

Where, (h, k) is the vertex.

So, let us convert the given equation in parabolic equation.

$\begin{array}{l}{y=-x^{2}+12 x-4} \\ {y=-\left(x^{2}-12 x+4\right)} \\ {y=-\left(x^{2}-(2)(x)(6)+6^{2}-6^{2}+4\right)} \\ {\left.y=-\left((x-6)^{2}-36+4\right)\right)} \\ {y=-(x-6)^{2}+32}\end{array}$

Now, by comparing above equation with general form,  

h = 6, k = 32 and a = -1.

Hence, the vertex of given equation is (6, 32).