# How do you find the vertex and intercepts for y² - 4y - 8x +25 = 0?

Dec 30, 2015

Vertex is $\left(\frac{21}{8} , 2\right)$, x intercept is $x = 25$ and there are no y intercepts.

#### Explanation:

First reorder to get one of the variables (in this case x) on its own.$8 x = {y}^{2} - 4 y + 25$ or $x = {y}^{2} / 8 - \frac{y}{2} + \frac{25}{8}$

Then use completing the squares to get the expression into the form $m {\left(x + \frac{b}{2}\right)}^{2} + n$. This is easiest done before dividing across the board by 8.
$8 x = {\left(y - 2\right)}^{2} - 4 + 25 = {\left(y - 2\right)}^{2} + 21$
$x = {\left(y - 2\right)}^{2} / 8 + \frac{21}{8}$

The vertex occurs where the bracketed term is zero and is therefore $\left(\frac{21}{8} , 2\right)$
The x intercept occurs where $y = 0$ and is therefore $x = 25$
The y intercepts occur where $x = 0$ and in this case it can be established by eye that the graph will never cross the y axis. Alternatively this can be shown using the quadratic formula.
$y = \frac{- \left(- \frac{1}{2}\right) \pm \sqrt{{\left(\frac{1}{2}\right)}^{2} - 4 \cdot 1 \cdot \left(\frac{25}{8}\right)}}{2 \cdot 1}$
$y = \frac{\frac{1}{2} \pm \sqrt{\frac{1}{4} - \frac{25}{2}}}{2} = \frac{\frac{1}{2} \pm \frac{\sqrt{2 - 25}}{2}}{2} = \frac{1 \pm \sqrt{- 23}}{4}$
This shows that there are no real roots to the equation, only imaginary ones.