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

Apr 20, 2018

Vertex is at $\left(0.5 , 2.5\right)$, y intercept is $y = 2 \mathmr{and} \left(0 , 2\right)$ and
x intercepts are at
$\left(- 0.618 , 0\right) \mathmr{and} \left(1.618 , 0\right)$

Explanation:

y =-2 x^2+2 x+2 ; a= -2 , b= 2 ,c =2

Vertex ( x coordinate) is ${v}_{x} = - \frac{b}{2 a} = \frac{- 2}{-} 4 = \frac{1}{2} = 0.5$

Putting $x = 2$ in the equation we get ${v}_{y}$

Vertex ( y coordinate) is ${v}_{y} = - 2 \cdot \frac{1}{4} + 2 \cdot \frac{1}{2} + 2$

$= - \frac{1}{2} + 1 + 2 = 2.5 \therefore$ Vertex is at $\left(0.5 , 2.5\right)$

y intercept is found by putting $x = 0$ in the equation

y = -2 x^2+2 x+2:. y=2 ; y intercept is $y = 2 \mathmr{and} \left(0 , 2\right)$

x intercept is found by putting $y = 0$ in the equation

$y = - 2 {x}^{2} + 2 x + 2 \mathmr{and} - 2 {x}^{2} + 2 x + 2 = 0$ or

$- 2 \left({x}^{2} + x\right) = - 2 \mathmr{and} - 2 {\left(x + 0.5\right)}^{2} = - 2 - 0.5$

$- 2 {\left(x - 0.5\right)}^{2} = - 2.5 \mathmr{and} {\left(x - 0.5\right)}^{2} = 1.25$ or

$\left(x - 0.5\right) = \pm \sqrt{1.25} \therefore x = 0.5 \pm 1.118$

$\therefore x \approx 1.618 \mathmr{and} x \approx - 0.618$

x intercepts are at $\left(- 0.618 , 0\right) \mathmr{and} \left(1.618 , 0\right)$

graph{-2x^2+2x+2 [-10, 10, -5, 5]} [Ans]