Answer:
[tex]\large\boxed{1.\ f(x)=(x-6)^2+10}\\\\\boxed{2.\ x=-\dfrac{3}{2}}[/tex]
Step-by-step explanation:
The quadratic function:
[tex]f(x)=ax^2+bx+c[/tex]
1.
The vertex form of a quadratic function:
[tex]f(x)=a(x-h)^2+k[/tex]
[tex]h=\dfrac{-b}{2a}\\\\k=f(h)[/tex]
We have
[tex]f(x)=x^2-12x+46\\\\a=1,\ b=-12,\ c=46[/tex]
Substitute:
[tex]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[/tex]
2.
The equation of an axis of symmetry:
[tex]x=\dfrac{-b}{2a}[/tex]
We have:
[tex]f(x)=3x^2+9x+15\\\\a=3,\ b=9,\ c=15[/tex]
Substitute:
[tex]x=\dfrac{-9}{2(3)}=-\dfrac{3}{2}[/tex]
