Question

f(x)=2x^2-12x+16
x intercept(s)
axis of symmetry

Answer 1

any idea?

Answer 2

\[f(x)=2x^2-12x+16\]
\[=2(x^2-6x+9-9)+16\]
\[=2(x-3)^2-18+16\]
\[=2(x-3)^2-2\]
vertex (3,-2)
axis of symmetry x-3=0
or x=3
to find x-intercepts put f(x)=0
\[2x^2-12x+16=0\]
or
\[x^2-6x+8=0\]
\[x^2-2x-4x+8=0\]
\[x(x-2)-4(x-2)=0\]
\[(x-2)(x-4)=0\]
x=2,4
x intercepts are (2,0),(4,0)

Answer 3

yw