what is $h(x)=-3x^{2}-6x + 5$ written in vertex form?\n$h(x)=-3(x + 1)^{2}+2$\n$h(x)=-3(x +…

what is $h(x)=-3x^{2}-6x + 5$ written in vertex form?\n$h(x)=-3(x + 1)^{2}+2$\n$h(x)=-3(x + 1)^{2}+8$\n$h(x)=-3(x - 3)^{2}-4$\n$h(x)=-3(x - 3)^{2}+32$

what is $h(x)=-3x^{2}-6x + 5$ written in vertex form?\n$h(x)=-3(x + 1)^{2}+2$\n$h(x)=-3(x + 1)^{2}+8$\n$h(x)=-3(x - 3)^{2}-4$\n$h(x)=-3(x - 3)^{2}+32$

Answer

Explanation:

Step1: Factor out -3 from first two terms

[ \begin{align*} h(x)&=-3(x^{2}+2x)+5 \end{align*} ]

Step2: Complete the square inside the parentheses

For the quadratic expression (x^{2}+2x), we know that ((x + m)^2=x^{2}+2mx+m^{2}), here (2m = 2) (so (m = 1)) and (x^{2}+2x=(x + 1)^2-1). Then (h(x)=-3((x + 1)^2-1)+5).

Step3: Expand and simplify

[ \begin{align*} h(x)&=-3(x + 1)^2+3 + 5\ h(x)&=-3(x + 1)^2+8 \end{align*} ]

Answer:

(h(x)=-3(x + 1)^2+8) (the second option)