Question

Y=3/4x^2+9x vertex form?

Answer 1

The standard form:

$f(x)=ax^2+bx+c$

The vertex form:

$f(x)=a(x-h)^2+k\\\\h=\dfrac{-b}{2a};\ k=f(h)$

We have

$y=\dfrac{3}{4}x^2+9x\to f(x)=\dfrac{3}{4}x^2+9x\\\\a=\dfrac{3}{4},\ b=9,\ c=0\\\\h=\dfrac{-b}{2a}\to h=\dfrac{-9}{2\cdot\frac{3}{4}}=\dfrac{-9}{\frac{3}{2}}=-9\cdot\dfrac{2}{3}=-3\cdot2=-6\\\\k=f(h)\to k=f(6)=\dfrac{3}{4}(-6)^2+9(-6)=\dfrac{3}{4}(36)-54\\=(3)(9)-54=27-54=-27$

Answer:

$y=\dfrac{3}{4}(x-(-6))^2+(-27)=\dfrac{3}{4}(x+6)^2-27$