How do you find the vertex and intercepts for #y=-3(x-3)(x+1)#?

#"to find the intercepts, that is where the graph crosses"#
#"the x and y axes"#

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

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

#x=0toy=-3(-3)(1)=9larrcolor(red)" y-intercept"#

#y=0to-3(x-3)(x+1)=0#

#"equate each factor to zero and solve for x"#

#x-3=0rArrx=3#

#x+1=0rArrx=-1#

#rArrx=-1,x=3larrcolor(red)" x-intercepts"#

#color(blue)"the axis of symmetry"" is at the midpoint of the"#
#"x-intercepts"#

#rArrx=(-1+3)/2=1rArrx=1" is the axis of symmetry"#

#"the vertex lies on the axis of symmetry"#

#rArr" x-coordinate of vertex is "x=1#

#"substitute this value into the equation for y-coordinate"#

#rArry_(color(red)"vertex")=-3(-2)(2)=12#

#rArrcolor(magenta)"vertex "=(1,12)#
graph{-3(x-3)(x+1) [-40, 40, -20, 20]}

#y= -3(x-3)(x+1) or y= -3(x^2-2x-3)# or

#y= -3(x^2-2x) +9 or y= -3(x^2-2x+1)+3+9# or

#y= -3(x-1)^2+12#. Comparing with vertex form of equation

#y=a(x-h)^2+k ; (h,k)# being vertex we find here

#h=1,k=12 :.# Vertex is at # (1,12)#

x intercepts can be found by putting #y=0# in the equation

#:. y= -3(x-3)(x+1) or -3(x-3)(x+1)=0# or

#(x-3)(x+1)=0 :.(x-3)=0 or x=3 ,(x+1)=0# or

#x= -1 :. # x intercepts are at # (-1,0) and (3,0)#

y intercept can be found by putting #x=0# in the equation

#:. y= -3(x-3)(x+1) or y=-3(0-3)(0+1)=9 :.#

y intercept is at #(0,9)#

graph{-3(x-3)(x+1) [-40, 40, -20, 20]} [Ans]