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"#