# How do you find the vertex and intercepts for  f(x) = -9x^2 + 7x + 9?

Nov 7, 2017

Vertex is $\left(\frac{7}{18} , 13 \frac{13}{36}\right)$, zeroes are approximately $- .684 \mathmr{and} 1.462$

#### Explanation:

If you write a quadratic $f \left(x\right) = a {x}^{2} + b x + c$, you can find the $x$ coordinate of the vertex using the formula $\frac{- b}{2 a}$. Let's plug our numbers into this to find the $x$.

$\text{-7"/"2(-9)" rArr "-7"/"-18} = \frac{7}{18}$

Next, we plug $x$ into $f \left(x\right)$ to find the $y$ of our vertex.

$f \left(\frac{7}{18}\right) = - 9 {\left(\frac{7}{18}\right)}^{2} + 7 \left(\frac{7}{18}\right) + 9$
$f \left(\frac{7}{18}\right) = 13 \frac{13}{36}$

So your vertex is $\left(\frac{7}{18} , 13 \frac{13}{36}\right)$

${x}_{1 , 2} = \frac{- b \pm \sqrt{\textcolor{b l u e}{{b}^{2} - 4 a c}}}{2 a}$

Let's first solve our discriminate, the numbers marked in blue above.

$\textcolor{b l u e}{{b}^{2} - 4 a c} \Rightarrow {\left(7\right)}^{2} - 4 \left(- 9\right) \left(9\right) = 373$

Now we can plug this into our quadratic formula, along with the other numbers.

${x}_{1 , 2} = \frac{- \left(7\right) \pm \sqrt{373}}{2 \left(- 9\right)}$

${x}_{1} = \frac{- 7 + \sqrt{373}}{-} 18 \approx - .684$

${x}_{2} = \frac{- 7 - \sqrt{373}}{-} 18 \approx 1.462$