# How do you find the axis of symmetry, and the maximum or minimum value of the function y=x^2+3x-5?

Apr 23, 2018

The axis of symmetry is $x = - \frac{3}{2}$ or $- 1.5$.

The vertex is $\left(- \frac{3}{2} , - \frac{29}{4}\right)$ or $\left(- 1.5 , - 7.25\right)$.

#### Explanation:

Given:

$y = {x}^{2} + 3 x - 5$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 1$, $b = 3$, $c = - 5$

Axis of symmetry: vertical line that divides the parabola into two equal halves. It is also the $x$-coordinate of the vertex.

The formula to find the axis of symmetry:

$x = \frac{- b}{2 a}$

Plug in the known values.

$x = \frac{- 3}{2 \cdot 1}$

$x = - \frac{3}{2}$ or $- 1.5$

The axis of symmetry is $x = - \frac{3}{2}$ or $- 1.5$.

Vertex: the minimum or maximum point on the parabola. The axis of symmetry is the $x$-coordinate. To find the $y$-coordinate, substitute $- \frac{3}{2}$ into the equation and solve for $y$.

$y = {\left(- \frac{3}{2}\right)}^{2} + 3 \left(- \frac{3}{2}\right) - 5$

$y = \frac{9}{4} - \frac{9}{2} - 5$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$, and $5$ by $\frac{4}{4}$ so each term has $4$ as its denominator.

$y = \frac{9}{4} - \frac{9}{2} \times \frac{\textcolor{red}{2}}{\textcolor{red}{2}} - 5 \times \frac{\textcolor{g r e e n}{4}}{\textcolor{g r e e n}{4}}$

$y = \frac{9}{4} - \frac{18}{4} - \frac{20}{4}$

Simplify.

$y = \frac{\left(9 - 18 - 20\right)}{4}$

$y = - \frac{29}{4}$

The vertex is $\left(- \frac{3}{2} , - \frac{29}{4}\right)$ or $\left(- 1.5 , - 7.25\right)$.

graph{y=x^2+3x-5 [-16.02, 16.01, -8.01, 8.01]}