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=(-b)/(2a);\ k=f(h)`

We have

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

Answer:

`y=(3)/(4)(x-(-6))^2+(-27)=(3)/(4)(x+6)^2-27`