Question 138711
The y-intercept occurs where {{{x=0}}}
{{{f(x) = 16x^2 + 40x + 25}}}
Make {{{x=0}}}
{{{f(0) = 25}}}
(0,25) is the y-intercept
The x-intercept occurs where {{{y=0}}}
{{{16x^2 + 40x + 25 = 0}}}
divide both sides by {{{16}}}
{{{x^2 + (5/2)x + (5/4)^2 = 0}}}
The left side is a perfect square
{{{(x + (5/4))^2 = 0}}}
Take the square root of both sides
{{{+(x + (5/4)) = 0}}}
{{{x = -(5/4)}}} 
and
{{{-(x + (5/4)) = 0}}}
{{{x = -(5/4)}}} 
So, the x-intercept is (-5/4,0)