Question

Please answer!!

Answer 1

Answer:

See below

Step-by-step explanation:

$f(x)=3x^2+6x-24\\\\f(x)=3(x^2+2x-8)\\\\f(x)+3(9)=3(x^2+2x-8+9)\\\\f(x)+27=3(x^2+2x+1)\\\\f(x)=3(x+1)^2-27$

Here, we can see that the vertex of the parabola is $(-1,-27)$ when we compare to the equation $y=(x-h)^2+k$ with vertex $(h,k)$.

A second point you could plot is the y-intercept, when $x=0$:

$f(0)=3(0+1)^2-27\\\\f(0)=3-27\\\\f(0)=-24$

So, your y-intercept is $(0,-24)$.