# How do you find the vertex and intercepts for g(x) = x^2 - 9x + 2?

Jan 15, 2018

Vertex: $\left(4.5 , - 18.25\right)$
x-intercepts: $\left(\frac{9 + \sqrt{73}}{2} , 0\right)$ and $\left(\frac{9 - \sqrt{73}}{2} , 0\right)$
y-intercept: $\left(0 , 2\right)$

#### Explanation:

$g \left(x\right) = {x}^{2} - 9 x + 2$

This equation is in standard form, or $y = a {x}^{2} + b x + c$.

The vertex is the highest or lowest point on the graph, depending on the coefficient (value before the ${x}^{2}$ if any).

To find the vertex, we first find the $x$-value of the vertex using this equation: $x = \frac{- b}{2 a}$.

In our equation, we know that $a = 1$ and $b = - 9$, so let's plug them into the equation:
$x = \frac{- \left(- 9\right)}{2 \left(1\right)}$
$x = \frac{9}{2}$ or $4.5$

Now let's find the $y$-value of the vertex. To do so, we plug in what we got for $x$ into the original equation:
$g \left(x\right) = {\left(4.5\right)}^{2} - 9 \left(4.5\right) + 2$
$g \left(x\right) = 20.25 - 40.5 + 2$
$g \left(x\right) = - 18.25$

So our vertex of the equation is $\left(4.5 , - 18.25\right)$.

To find the x-intercepts...
We plug in $0$ for the $y$-values in the equation and solve for $x$:
$0 = {x}^{2} - 9 x + 2$

Now, to solve this, we need to use the quadratic formula, which is this long equation :(
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$
$x = \frac{9 \pm \sqrt{81 - 8}}{2}$
$x = \frac{9 \pm \sqrt{73}}{2}$

To find the y-intercepts...
We plug in $0$ for the $x$-values in the equation and solve for $y$:
$g \left(x\right) = {0}^{2} - 9 \left(0\right) + 2$
$g \left(x\right) = 0 - 0 + 2$
$g \left(x\right) = 2$

So our y-intercept is at $\left(0 , 2\right)$.

To show that the vertex and intercepts are correct, here is a graph of this equation: If you need more help on this type of question, feel free to watch this video:

If you need more help on quadratic formula, feel free to watch this video:

Hope this helps!