# How do you find the vertex and intercepts for y = 10x – 3x^2?

Jun 17, 2016

The vertex is $\left(\frac{5}{3} , \frac{25}{3}\right)$
The intercepts are (0,0) and $\left(\frac{10}{3} , 0\right)$

#### Explanation:

you can find the coordinate x of the vertex using the formula
$x = - \frac{b}{2 a}$
where a is the coefficient of ${x}^{2}$ and b of x

Since a=-3 and b=10, you have

$x = - \frac{10}{-} 6 = \frac{5}{3}$

Then you can put this value in x and find y
$y = 10 \left(\frac{5}{3}\right) - 3 {\left(\frac{5}{3}\right)}^{2}$

$y = \frac{50}{3} - 3 \left(\frac{25}{9}\right)$
$y = \frac{50}{3} - \frac{25}{3}$
$y = \frac{25}{3}$

One of the intercepts is the origin (0,0) because y has no known term

By solving the equation $10 x - 3 {x}^{2} = 0$ you can find the other one:

$x \left(10 - 3 x\right) = 0$

$10 - 3 x = 0$

$x = \frac{10}{3}$