Question

Help please!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

$\large\boxed{1.\ f(x)=(x-6)^2+10}\\\\\boxed{2.\ x=-\dfrac{3}{2}}$

Step-by-step explanation:

The quadratic function:

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

1.

The vertex form of a quadratic function:

$f(x)=a(x-h)^2+k$

$h=\dfrac{-b}{2a}\\\\k=f(h)$

We have

$f(x)=x^2-12x+46\\\\a=1,\ b=-12,\ c=46$

Substitute:

$h=\dfrac{-(-12)}{2(1)}=\dfrac{12}{2}=6\\\\k=f(6)=6^2-12(6)+46=36-72+46=10\\\\f(x)=1(x-6)^2+10$

2.

The equation of an axis of symmetry:

$x=\dfrac{-b}{2a}$

We have:

$f(x)=3x^2+9x+15\\\\a=3,\ b=9,\ c=15$

Substitute:

$x=\dfrac{-9}{2(3)}=-\dfrac{3}{2}$