# How do you find the vertex and intercepts for x^2-10x-8y+33=0?

Mar 11, 2017

${y}_{\text{intercept}} = \frac{33}{8}$

Vertex$\to \left(x , y\right) = \left(5 , 1\right)$

NO X-INTERCEPT

#### Explanation:

Moving $8 y$ to the other side of the = and change its sign

${x}^{2} - 10 x + 33 = + 8 y$

To get $y$ on its own divide both sides by 8

$\frac{{x}^{2}}{8} - \frac{10}{8} x + \frac{33}{8} = y$

Write as:

$y = \frac{{x}^{2}}{8} - \frac{10}{8} x + \frac{33}{8}$

Write as:

$y = \frac{1}{8} \left({x}^{2} \textcolor{red}{- 10} x\right) + \frac{33}{8}$

color(green)(y_("intercept")=+33/8 larr" read directly off the equation")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- 10}\right) = + 5}$
The above line is part of the process of completing the square.

$\textcolor{g r e e n}{\text{The "x^2/8"is positive so the graph is of general shape } \cup}$

Substituting $x = + 5$ gives:

$\textcolor{g r e e n}{{y}_{\text{vertex}} = \frac{1}{8} \left[\textcolor{w h i t e}{\frac{.}{.}} {5}^{2} - 10 \left(5\right) \textcolor{w h i t e}{.}\right] + \frac{33}{8} = 1}$

$\textcolor{g r e e n}{\text{Vertex} \to \left(x , y\right) = \left(5 , 1\right)}$

As the graph is of general shape $\cup$ and ${y}_{\text{vertex}}$ is above the x-axis there is NO X-INTERCEPT 