Question

Find the vertex of this parabola y=-x^2+8x-22

Answer 1

Vertex of parabola is y= $-x^{2} +8x-22$ at (4,6)

Step-by-step explanation:

The given equation of parabola is y= $-x^{2} +8x-22$

Simplifying the equation,

y= $-x^{2} +8x-22$

y= $(-1)(x^{2}-8x-+22)$

y= $(-1)(x^{2} -8x + 16-16+22)$

y= $(-1)[(x^{2} -8x + 16)-(16-22)]$

y= $(-1)[(x-4)^{2}+(+6)]$

y= $(-1)(x-4)^{2}+(+6)(-1)$

y= $(-1)(x-4)^{2}+(-6)$

The general equation of parabola is y = y= $a(x+h)^{2}+k$

Where, (h,k) is vertex of parabola.

On comparing the equations

we get,

Vertex of parabola is y= $-x^{2} +8x-22$ at (4,-6)