What is the axis of symmetry and vertex for the graph y = (x – 1)^2 – 16?

Jul 15, 2017

$x = 1 \text{ and } \left(1 , - 16\right)$

Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where ( h , k ) are the coordinates of the vertex and a is a constant.

$y = {\left(x - 1\right)}^{2} - 16 \text{ is in this form}$

$\text{with " h=1" and } k = - 16$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , - 16\right)$

$\text{since " a>0" this graph opens up vertically}$

$\text{the axis of symmetry passes through the vertex}$

$\text{with equation } x = 1$