How do you find the axis of symmetry, vertex and x intercepts for y=-2x^2-4x-1?

1 Answer
Dec 21, 2017

"see explanation"

Explanation:

"the equation of a parabola in "color(blue)"vertex form" is.

color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))

"where "(h,k)" are the coordinates of the vertex and a"
"is a multiplier"

"the equation of the axis of symmetry is "x=h

"to obtain this form use "color(blue)"completing the square"

• " the coefficient of the "x^2" term must be 1"

rArry=-2(x^2+2x)-1

• " add/subtract "(1/2"coefficient of x-term")^2" to"
x^2+2x

y=-2(x^2+2(1)xcolor(red)(+1)color(red)-1))-1

color(white)(y)=-2(x+1)^2+2-1

color(white)(y)=-2(x+1)^2+1larrcolor(red)"in vertex form"

rArrcolor(magenta)"vertex "=(-1,1)

"equation of axis of symmetry is "x=-1

"to find the x-intercepts set y = 0"

rArr-2(x+1)^2+1=0

rArr(x+1)^2=1/2

rArrx+1=+-1/2=+-1/sqrt2

rArrx=-1+-1/sqrt2larrcolor(red)"exact values"

rArrx~~-1.71,x~~-0.29larrcolor(red)"x-intercepts"
graph{(y+2x^2+4x+1)(y-1000x-1000)=0 [-10, 10, -5, 5]}