Question

What is the vertex of the parabola using this equation y=-x^2+4x+5

Answer 1

(2, 9)

Explanation:

`y=-x^2+4x+5\quad \implies\quad a=-1\,,\ b=4\,,\ c=5\\\\h=(-b)/(2a)=(-4)/(2(-1))=2\\\\k=c-(b^2)/(4a)=5-(4^2)/(4(-1))=5-(16)/(-4)=5+4=9`

Or by completing the square:

`y=-x^2+4x+5\\\\y=-(x^2-4x)+5\\\\y=-(\underline{x^2-2\cdot x\cdot 2+2^2}-2^2)+5\\\\y=-\left((x-2)^2-4\right)+5\\\\y=-(x-2)^2+4+5\\\\y=-(x-2)^2+9\quad\implies\quad h=2\,,\ k=9`