Answer
The graph of $y=4x^2+4x+1$ has one $x-$intercept.
Work Step by Step
The given function is
$\Rightarrow y=4x^2+4x+1$
Find the number of real solutions of $0=4x^2+4x+1$
$= b^2-4ac$
Substitute $4$ for $a,4$ for $b,$ and $1$ for $c$.
$= (4)^2-4(4)(1)$
Simplify.
$= 16-16$
Subtract.
$= 0$
Discriminant is equal to $0$,
Hence, the equation has one real solution.
The graph of $y=4x^2+4x+1$ has one $x-$intercept.
