# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y= x^2+ 5x - 12?

Jun 14, 2015

$y = {x}^{2} + 5 x - 12 = {\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} - 12$

$= {\left(x + \frac{5}{2}\right)}^{2} - \frac{73}{4}$

Hence vertex $\left(- \frac{5}{2} , - \frac{73}{4}\right)$, axis of symmetry $x = - \frac{5}{2}$, $y$ intercept $\left(0 , - 12\right)$, $x$ intercepts $\left(- \frac{5}{2} \pm \frac{\sqrt{73}}{2} , 0\right)$

#### Explanation:

${\left(x + \frac{5}{2}\right)}^{2} = {x}^{2} + 5 x + \frac{25}{4}$

So

$y = {x}^{2} + 5 x - 12 = {\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} - 12$

$= {\left(x + \frac{5}{2}\right)}^{2} - \frac{73}{4}$

The minimum value of ${\left(x + \frac{5}{2}\right)}^{2}$ occurs when $\left(x + \frac{5}{2}\right) = 0$, that is when $x = - \frac{5}{2}$

If $x + \frac{5}{2} = 0$ then $y = - \frac{73}{4}$.

So the vertex is at $\left(- \frac{5}{2} , - \frac{73}{4}\right)$

The parabola is vertical, so the axis of symmetry is just the vertical line through the vertex, i.e. $x = - \frac{5}{2}$.

The intercept with the $y$ axis occurs where $x = 0$. From the original equation we find $y = {x}^{2} - 5 x - 12 = {0}^{2} - 5 \cdot 0 - 12 = - 12$. So the $y$ intercept is at $\left(0 , - 12\right)$.

The intercepts with the $x$ axis are where $y = 0$. Then:

${\left(x + \frac{5}{2}\right)}^{2} - \frac{73}{4} = 0$

So ${\left(x + \frac{5}{2}\right)}^{2} = \frac{73}{4}$

So $x + \frac{5}{2} = \pm \sqrt{\frac{73}{4}} = \pm \frac{\sqrt{73}}{2}$

and $x = - \frac{5}{2} \pm \frac{\sqrt{73}}{2}$

That is the $x$ intercepts are $\left(- \frac{5}{2} \pm \frac{\sqrt{73}}{2} , 0\right)$