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

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Answer 1 · 2018-04-20T12:53:18

Vertex is at $(0.5,2.5)$ , y intercept is $y=2 or (0,2)$ and
x intercepts are at $(-0.618,0) and (1.618,0)$

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

Vertex ( x coordinate) is $v_x= -b/(2 a)=(-2)/-4=1/2=0.5$

Putting $x=2$ in the equation we get $v_y$

Vertex ( y coordinate) is $v_y= -2 *1/4+2*1/2+2$

$= -1/2 + 1+2= 2.5 :.$ Vertex is at $(0.5,2.5)$

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

$y = -2 x^2+2 x+2:. y=2 ;$ y intercept is $y=2 or (0,2)$

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

$y = -2 x^2+2 x+2 or -2 x^2+2 x+ 2=0$ or

$-2( x^2+ x) = -2 or -2 ( x+0.5)^2 = -2- 0.5$

$-2(x-0.5)^2=-2.5 or (x-0.5)^2 = 1.25$ or

$(x-0.5) = +- sqrt 1.25 :. x = 0.5 +- 1.118$

$:. x ~~ 1.618 and x ~~ -0.618$

x intercepts are at $(-0.618,0) and (1.618,0)$

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

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