How do you find the vertex and the intercepts for #y = -3(x - 2)^2 + 4#?

1 Answer
Sep 17, 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.

#y=-3(x-2)^2+4" is in vertex form"#

#"with "(h,k)=(2,4)larrcolor(red)" vertex"#

#"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-3(x-2)^2=-4#

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

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

#rArrx-2=+-sqrt(4/3)larr" note plus or minus"#

#rArrx=2+-2/sqrt3=2+(2sqrt3)/3#

#rArrx~~ 0.85" or "x~~ 3.15larrcolor(red)" x-intercepts"#
graph{-3(x-2)^2+4 [-10, 10, -5, 5]}