Question

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

Answer 1

`(1,12),x=-1,x=3`

Explanation:

`"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]}

Answer 2

Vertex is at `(1,12)`, x intercepts are at `(-1,0) and (3,0)` and
y intercept is at `(0,9)`

Explanation:

`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]