# How do you find the axis of symmetry, and the maximum or minimum value of the function y= -x^2-10x+7?

Jul 21, 2018

$x = - 5 \text{ and maximum value } = 32$

#### Explanation:

$\text{We require to find the vertex and determine if maximum}$
$\text{or minimum turning point}$

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

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$x = h \text{ is the axis of symmetry}$

$\text{to obtain this form use "color(blue)"completing the square}$

$y = - \left({x}^{2} + 10 x - 7\right)$

$\textcolor{w h i t e}{y} = - \left({x}^{2} + 2 \left(5\right) x + 25 - 25 - 7\right)$

$\textcolor{w h i t e}{y} = - {\left(x + 5\right)}^{2} + 32$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(- 5 , 32\right)$

$\text{Since "a<0" then maximum turning point } \bigcap$

$\text{axis of symmetry is } x = - 5$

$\text{and maximum value } = 32$
graph{-x^2-10x+7 [-80, 80, -40, 40]}

Axis of symmetry: $x + 5 = 0$

Maximum value $= 32$

#### Explanation:

The given equation:

$y = - {x}^{2} - 10 x + 7$

$y = - \left({x}^{2} + 10 x + 25\right) + 25 + 7$

$y = - {\left(x + 5\right)}^{2} + 32$

${\left(x + 5\right)}^{2} = - \left(y - 32\right)$

The above equation is in standard form of downward parabola ${\left(x - {x}_{1}\right)}^{2} = - 4 a \left(y - {y}_{1}\right)$ which has

Axis of symmetry: $x - {x}_{1} = 0$

$x + 5 = 0$

Vertex $\left(x - {x}_{1} = 0 , y - {y}_{1} = 0\right)$

$\left(x + 5 = 0 , y - 32 = 0\right) \setminus \equiv \left(- 5 , 32\right)$

Maximum value of given quadratic function will be at the vertex $\left(- 5 , 32\right)$ of given downward parabola: $y = - {x}^{2} - 10 x + 7$

Hence, the maximum value of given function is obtained by setting $x = - 5$ in the function as follows

$y \left(- 5\right) = - {\left(- 5\right)}^{2} - 10 \left(- 5\right) + 7$

$= 32$