# How do you find the zeros by rewriting the function y=x^2+16x+64 in intercept form?

Aug 27, 2017

$y = {\left(x + 8\right)}^{2}$

#### Explanation:

We write the function as $y = f \left(x\right)$ and factorize $f \left(x\right)$ to find intercepts on $x$-axis. If $f \left(x\right) = a \left(x - \alpha\right) \left(x - \beta\right)$, the zeros of $f \left(x\right)$ are $\alpha$ and $\beta$ and intercepts are $\alpha$ and $\beta$ and $y = a \left(x - \alpha\right) \left(x - \beta\right)$ is the equation in intercept form.

Here we have $y = {x}^{2} + 16 x + 64 = {\left(x + 8\right)}^{2}$

Hence, we have just one intercept $x = - 8$

graph{(x+8)^2 [-16.42, 3.58, -4.08, 5.92]}