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Find the $y$ -intercept and any $x$ -intercept(s) of the quadratic curve.

${y}={x^{2}+4x+4}$

If the parabola does not have any $x$ -intercept, type DNE, meaning does not exist. If the parabola has two $x$ -intercepts, use a comma to separate them.

$y$ -intercept:

$x$ -intercept(s):

Solution

To find a parabola’s $y$ -intercept, we substitute $x=0$ into ${y}={x^{2}+4x+4}$ , and we have $y=4$ .

The parabola’s $y$ -intercept is $(0,4)$ .

To find a parabola’s $x$ -intercept, we substitute $y=0$ into ${y}={x^{2}+4x+4}$ , and then solve for $x$ .

For this problem, the polynomial can be factored:

$\displaystyle{\begin{aligned}[t] y &= {x^{2}+4x+4} \\ 0 &= {x^{2}+4x+4} \\ 0 &= (x + 2)(x + 2) \\ x + 2 &= 0 \\ x &= -2 \\ \end{aligned} }$

The parabola has only one $x$ -intercept: $(-2,0)$ .