# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x^2-7x-28?

Jun 9, 2015

Find vertex of y = x^2 - 7x - 28

#### Explanation:

x of vertex: $x = \left(- \frac{b}{2} a\right) = \frac{7}{2}$

y of vertex: $y = f \left(\frac{7}{2}\right) = \frac{49}{4} - \frac{49}{2} - 28 = \frac{161}{4}$

x of axis of symmetry = x of vertex =$\frac{7}{2}$

To find y-intercept, make x = 0 -> y = -28
To find x-intercepts, make y = 0
$D = {d}^{2} = {b}^{2} - 4 a c$ = 49 + 112 = 161 -->$d = \pm \sqrt{161}$

x-intercepts:$x = \frac{7}{2} \pm \frac{\sqrt{161}}{2}$