# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x^2+3x-5?

Mar 7, 2018

$\text{axis of symmetry is } x = - \frac{3}{2}$
${x}_{\text{intercept")~~-4.19 and x_("intercept}} \approx 1.19$ to 2 decimal places
${y}_{\text{intercept}} = - 5$

#### Explanation:

You have several approaches available.

$\textcolor{b l u e}{\text{Axis of symmetry - a sort of cheat way}}$

Let me show you a quick way of determining the axis of symmetry.

Standardised form of $y = a {x}^{2} + b x + c$

Write as

$y = a \left({x}^{2} + \frac{b}{a} x\right) + c \textcolor{w h i t e}{\text{ddd") ->color(white)("ddd}} y = 1 \left({x}^{2} + \frac{3}{1} x\right) - 5$
This is the beginnings if completing the square

x_("vertex")="axis of symmetry" = (-1/2)xxb/a

color(magenta)(x_("vertex")="axis of symmetry" = (-1/2)xx3=-3/2)

Another way is the determine the x-intercepts (if there is any) the axis of symmetry is 1/2 way between those 2 points.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the x-intercepts}}$

$y = {x}^{2} + 3 x - 5$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \to x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \left(1\right)}$

$x = - \frac{3}{2} \pm \frac{\sqrt{29}}{2}$

$x = - 4.19258 \ldots . \mathmr{and} x = + 1.19258 \ldots$

${x}_{1} \approx - 4.19 \mathmr{and} {x}_{2} \approx 1.19$ to 2 decimal places

Just to check the axis of symmetry

$\textcolor{m a \ge n t a}{\frac{{x}_{1} + {x}_{2}}{2} = - 1.5 = - \frac{3}{2} \leftarrow \text{ confirmed}}$
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$\textcolor{b l u e}{\text{Determine the y-intercepts}}$

This occures at $x = 0$

${y}_{\text{intercept}} = {0}^{2} + 3 \left(0\right) - 5 = - 5$