# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of y= -x^2-8x+10?

Oct 7, 2015

Vertex: (-4,26))
Axis of symmetry: $x = - 4$
y-intercept: $10$
x-intercepts: $- 4 \pm \sqrt{26}$
Domain: $\mathbb{R}$
Range: $\mathbb{R} \cap \left(- \infty , 26\right]$

#### Explanation:

Vertex
Re-writing $y = - {x}^{2} - 8 x + 10$ in vertex form:
color(white)("XXX")y= (-1)(x^2+8x+4^2)+10+16)
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 1\right) {\left(x - \left(\textcolor{red}{- 4}\right)\right)}^{2} + \left(\textcolor{b l u e}{26}\right)$
with vertex at $\left(\textcolor{red}{- 4} , \textcolor{b l u e}{26}\right)$

Axis of Symmetry
Any parabola in the form $y = a {x}^{2} + b x + c$ has as an axis of symmetry the vertical line passing through the vertex.

In this case $x = - 4$

y-intercept
The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = - {\left(0\right)}^{2} - 8 \left(0\right) + 10 = 10$

x-intercepts
The x-intercepts are the values of $x$ when $y = 0$
$0 = \left(- 1\right) {\left(x + 4\right)}^{2} + 26$
${\left(x + 4\right)}^{2} = 26$
$x + 4 = \pm \sqrt{26}$
$x = - 4 \pm \sqrt{26}$

Domain
$y = - {x}^{2} - 8 x + 10$ is defined for all Real values of $x$

Range
Since a parabolic equation with a negative coefficient for the ${x}^{2}$ term opens downward, the vertex gives a maximum value for the Range.
The Range is $\le 26$