# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x)=x^2-5x?

Jun 6, 2015

In the general case $f \left(x\right) = a {x}^{2} + b x + c$ we can complete the square as follows:

$f \left(x\right) = a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

The vertex of this parabola is where $x + \frac{b}{2 a} = 0$, that is when $x = - \frac{b}{2 a}$ and $y = c - {b}^{2} / \left(4 a\right)$.

That is $\left(- \frac{b}{2 a} , c - {b}^{2} / \left(4 a\right)\right)$

The axis of symmetry is the vertical line given by the equation

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

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

The intercepts with the $x$ axis are the roots of $f \left(x\right) = 0$ which can be calculated in general as:

$\left(\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} , 0\right)$

If $c = 0$ this simplifies to:

$\left(\frac{- b \pm b}{2 a} , 0\right)$

That is $\left(- \frac{b}{a} , 0\right)$ and $\left(0 , 0\right)$

In the case of $f \left(x\right) = {x}^{2} - 5 x$, we have $a = 1$, $b = - 5$ and $c = 0$.

The vertex is at $\left(- \frac{b}{2 a} , c - {b}^{2} / \left(4 a\right)\right) = \left(\frac{5}{2} , - \frac{25}{4}\right)$

The axis of symmetry is $x = - \frac{b}{2 a} = \frac{5}{2}$

The intercept with the $y$ axis is $\left(0 , c\right) = \left(0 , 0\right)$

Because $c = 0$, the intercepts with the $x$ axis are

$\left(- \frac{b}{a} , 0\right) = \left(5 , 0\right)$ and $\left(0 , 0\right)$