Question 62480
{{{f(x)=3(x-4)^2-1}}}

The axis of symmetry is at x=4 because x=4 makes the part in the bracket = 0.
The vertex is on the axis of symmetry so it will be at -1. Can either just look at the -1 at the end of the equation or substitute x=4 into the equation.

The y intercept occurs when x=0 so substitute x=0 into the equation.
{{{f(x)=3(0-4)^2-1}}}
{{{f(x)=3*16-1}}}
{{{f(x)=48-1}}}
{{{f(x)=47}}}

The x intercepts occur when f(x)=0 so substitute that into the equation.
{{{0=3(x-4)^2-1}}} Add 1 to each side of the equation
{{{0+1=3(x-4)^2-1+1}}}
{{{1=3(x-4)^2}}} Divide each side by 3
{{{1/3=3(x-4)^2/3}}}
{{{1/3=(x-4)^2}}} Find the square root of each side
{{{1/sqrt(3)=x-4}}} or {{{-(1/sqrt(3))=x-4}}} Solve for x by adding 4 to each side of the equation
{{{x = 4 + 1/sqrt(3)}}} or {{{x=4-(1/sqrt(3))}}}