How do you find the vertex and the intercepts for y= -3x^2 + 12x - 8?

1 Answer
Feb 8, 2018

"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"

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

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

rArry=-3(x^2-4x+8/3)

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
x^2-4x

y=-3(x^2+2(-2)xcolor(red)(+4)color(red)(-4)+8/3)

color(white)(y)=-3(x-2)^2-3(-4+8/3)

rArry=-3(x-2)^2+4larrcolor(red)"in vertex form"

rArrcolor(magenta)"vertex "=(2,4)

"to find the intercepts"

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

x=0toy=-3(-2)^2+4=-8larrcolor(red)"y-intercept"

y=0to-3(x-2)^2+4=0

rArr(x-2)^2=4/3

color(blue)"take square root of both sides"

x-2=+-sqrt(4/3)larrcolor(blue)"note plus or minus"

rArrx=2+-2/sqrt3

rArrx=2+-(2sqrt3)/3larrcolor(red)"x-intercepts"