How do you find the vertex and the intercepts for $f(x)=-2x^2+2x+4$ ?

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Answer 1 · 2017-11-25T06:50:17

Vertex $(1/2, 9/2)$
Y- intercept $(0, 4)$
X - intercept $(-1,0)$
X - intercept $(2,0)$

Given -

$f(x)=-2x^2+2x+4$

$y=-2x^2+2x+4$

Vertex

$x=(-b)/(2a)=(-2)/(2xx(-2))=(-2)/(-4)=1/2$

At $x=1/2$

$y=-2xx(1/2)^2+(2xx1/2)+4$

$y=-2xx(1/4)+1+4$

$y=--1/2+1+4=(-1+2+8)/2=9/2$

Vertex $(1/2, 9/2)$

Y- intercept
At $x=0$

$y=-2(0)^2+2(0)+4$

$y=4$

Y- intercept $(0, 4)$

X- intercept

At $y=0$

$-2x^2+2x+4=0$

$-2x^2+4x-2x+4=0$

$-2x(x-2)-2(x-2)=0$

$(-2x-2)(x-2)=0$

$-2x-2=0$
$x=(2)/(-2)=-1$

X - intercept $(-1,0)$

$x-2=0$

$x=2$

X - intercept $(2,0)$

How do you find the vertex and the intercepts for f(x)=-2x^2+2x+4f(x)=-2x^2+2x+4?