# How do you find the VERTEX of a parabola f(x)= -x^2 + 3x + 10?

Jul 17, 2015

Find vertex of$f \left(x\right) = - {x}^{2} + 3 x + 10$
Vertex $\left(\frac{3}{2} , \frac{49}{4}\right)$

#### Explanation:

x coordinate of vertex: $x = \left(- \frac{b}{2} a\right) = - \frac{3}{-} 2 = \frac{3}{2}$

y coordinate of vertex: $y = f \left(\frac{3}{2}\right) = - \frac{9}{4} + \frac{9}{2} + 10 = \frac{49}{4}$

Jul 17, 2015

$x = \frac{3}{2} \mathmr{and} 1 \frac{1}{2}$

$y = \frac{49}{4} \mathmr{and} 12 \frac{1}{4}$

The vertex is $\left(\frac{3}{2} , \frac{49}{4}\right)$ or $\left(1 \frac{1}{2} , 12 \frac{1}{4}\right)$.

#### Explanation:

$f \left(x\right) = - {x}^{2} + 3 x + 10$

Substitute $y$ for $f \left(x\right)$.

$y = - {x}^{2} + 3 x + 10$ is in the form $a {x}^{2} + b x + c$, where $a = - 1 , b = 3 , c = 10$.

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

$x = \frac{- 3}{2 \cdot - 1}$

$x = \frac{- 3}{- 2}$ =

$x = \frac{3}{2} = 1 \frac{1}{2}$

To find the $y$ value, substitute $\frac{3}{2}$ for $x$ in the equation $y = {x}^{2} + 3 x + 10$.

$y = - {\left(\frac{3}{2}\right)}^{2} + 3 \left(\frac{3}{2}\right) + 10$ =

$y = - \frac{9}{4} + \frac{9}{2} + 10$

The common denominator is $4$. Multiply the terms on the right side by the fraction that will give each term a denominator of $4$.

$y = - \frac{9}{4} + \frac{9}{2} \cdot \frac{2}{2} + 10 \cdot \frac{4}{4}$

$y = - \frac{9}{4} + \frac{18}{4} + \frac{40}{4}$ =

$y = \frac{49}{4} = 12 \frac{1}{4}$.