Question 196165
Any value I pick for {{{x}}} except {{{x = 4}}}
will give me a value for {{{f(x)}}} which is
greater than {{{-1}}} by some amount, so
(4, -1) is a minimum vertex.
The x-intercept is where {{{f(x)=0}}}
{{{f(x)= (x-4)^2-1}}}
{{{0 = (x-4)^2 - 1}}}
{{{x=5}}}
{{{x=3}}}
The x-intercepts are (5,0) and (3,0)
The y-intercept is where {{{x=0}}}
{{{f(x)= (x-4)^2-1}}}
{{{f(x)= (0-4)^2-1}}}
{{{f(x)= 15}}}
The y-intercept is at (0,15)
Here's the plot:
{{{ graph( 500, 500, -2, 10, -2, 20, (x-4)^2 - 1) }}}