# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x^2-x-4?

Jun 6, 2015

$y = {x}^{2} - x - 4 = {\left(x - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 4$

$= {\left(x - \frac{1}{2}\right)}^{2} - \frac{17}{4}$

The vertex is at $\left(\frac{1}{2} , - \frac{17}{4}\right)$

The axis of symmetry is the vertical line $x = \frac{1}{2}$

The intercept with the $y$ axis is $\left(0 , - 4\right)$

The intercepts with the $x$ axis are the points where $y = 0$, so

${\left(x - \frac{1}{2}\right)}^{2} = \frac{17}{4}$

$\left(x - \frac{1}{2}\right) = \pm \sqrt{\frac{17}{4}} = \pm \frac{\sqrt{17}}{2}$

So

$x = \frac{1}{2} \pm \frac{\sqrt{17}}{2} = \frac{1 \pm \sqrt{17}}{2}$

That is $\left(\frac{1 - \sqrt{17}}{2} , 0\right)$ and $\left(\frac{1 + \sqrt{17}}{2} , 0\right)$

Jun 6, 2015

x of vertex: $x = \left(- \frac{b}{2} a\right) = \frac{1}{2}$

y of vertex: $f \left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 4 = - \frac{17}{4}$

Axis of symmetry: x = (-b/2a) = 1/2

x-intercepts: y = x^2 - x - 4 = 0

$D = {d}^{2} = 1 + 16 = 19 \to d = \pm \sqrt{17}$

$x = \frac{1}{2} \pm \frac{\sqrt{17}}{2}$