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

##### 1 Answer

Jan 1, 2018

#### 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+3)^2" is in vertex form"#

#"with "h=-3" and "k=0#

#rArrcolor(magenta)"vertex "=(-3,0)#

#color(blue)"to find intercepts"#

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

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

#x=0toy=-3(9)=-27larrcolor(red)"y-intercept"#

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

#rArr(x+3)^2=0#

#rArrx+3=0rArrx=-3larrcolor(red)"x-intercept"#

graph{-3(x+3)^2 [-10, 10, -5, 5]}