# How do you find the maximum or minimum of y-x^2+6=9x?

Apr 27, 2018

The minimum is $\left(- 4.5 , - 26.25\right)$.

#### Explanation:

The maximum or minimum of a quadratic equation is the same as the vertex. First, we have to write the equation in standard form, or $y = a {x}^{2} + b x + c$

To do this, let's make $y$ by itself. Add ${x}^{2}$ and subtract $6$ from both sides of the equation:
$y - {x}^{2} + 6 \quad \textcolor{red}{+ \quad {x}^{2} \quad - \quad 6} = 9 x \quad \textcolor{red}{+ \quad {x}^{2} \quad - \quad 6}$

$y = {x}^{2} + 9 x - 6$

To find the $x$ value of the vertex, we use the formula $x = - \frac{b}{2 a}$.

Let's plug in the numbers:
$x = - \frac{9}{2 \left(1\right)}$

$x = - 4.5$

To find the $y$ value of the vertex, we substitute back in the value of $x$ back into the equation:
$y = {x}^{2} + 9 x - 6$

$y = {\left(- 4.5\right)}^{2} + 9 \left(- 4.5\right) - 6$

$y = 20.25 - 40.5 - 6$

$y = - 26.25$

Therefore, using the $x$ and $y$ values of the vertex, we know that the vertex is at $\left(- 4.5 , - 26.25\right)$. To verify this, let's graph it:

(desmos.com)

Because in this situation the vertex is the smallest point of the graph, we know that is is a minimum.

As you can see, the vertex/minimum is indeed $\left(- 4.5 , - 26.25\right)$.

Hope this helps!