# How do you find the vertex and the intercepts for y = -5x^2 + 10x - 7?

Feb 9, 2017

$V \left(1 : - 2\right)$

one intercept (0;-7)

#### Explanation:

You can find the vertex by applying the following formula:

${V}_{x} = - \frac{b}{2 a}$

and then substitute its value in the equation to find ${V}_{y}$:

#V_x=-10/(2(-5))=1

${V}_{y} = - 5 \cdot {1}^{2} + 10 \cdot 1 - 7 = - 2$

Then the vertex is $V \left(1 : - 2\right)$

To find the intercepts with y-axis, you would substitute $x = 0$ in the equation and, with x-axis, $y = 0$:

$x = 0 \to y = - 7$

$y = 0 \to - 5 {x}^{2} + 10 x - 7 = 0 \to x = \frac{- 5 \pm \sqrt{25 - 35}}{-} 5 \to$there are no real solutions
graph{-5x^2+10x-7 [-2,3, -10, 1]}