# How do you find the axis of symmetry, and the maximum or minimum value of the function y=x^2+5x-7?

Mar 14, 2017

$x = - \frac{5}{2} \text{ and minimum value } = - \frac{53}{4}$

#### Explanation:

For the standard equation of a parabola.

• y=ax^2+bx+c

• " has a minimum if "a>0" that is " uuu

• "has a maximum if "a<0" that is "nnn

$\text{for } y = {x}^{2} + 5 x - 7 , a = 1 , b = 5 , c = - 7$

$\text{Since "a>0" then parabola has a minimum}$

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a} = - \frac{5}{2}$

Substitute this value into the function for ${y}_{\text{vertex}}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = {\left(- \frac{5}{2}\right)}^{2} + \left(5 \times - \frac{5}{2}\right) - 7$

$\textcolor{w h i t e}{\text{rArry_(} v e} = \frac{25}{4} - \frac{50}{4} - \frac{28}{4}$

$\textcolor{w h i t e}{\times \times \times} = - \frac{53}{4}$

The vertex has coordinates $\left(- \frac{5}{2} , - \frac{53}{4}\right)$

The axis of symmetry passes through the vertex and is vertical.

$\Rightarrow \text{axis of symmetry is } x = - \frac{5}{2}$

$\text{and minimum value } = - \frac{53}{4}$
graph{(y-x^2-5x+7)(y-1000x-2500)=0 [-28.85, 28.89, -14.42, 14.44]}