Question

How do you find the vertex and the intercepts for `-3x^2 - 3x + 1`?

Answer 1

Vertex `((-1)/3, 5/3)`

Y-intercept

`(1, 0)`

X-Intercepts

`((3+sqrt(21))/(-6), 0)`
`((3-sqrt(21))/(-6), 0)`

Explanation:

Given -

`y= -3x^2-3x+1`

Find the vertex.

x- co-ordinate of the vetex

`x=(-(-b))/(2a)=(-(-3))/(2 xx (-3))=3/(-6)=(-1)/3`

Y-co-ordinate

`y=-3((-1)/3)-3((-1)/3)+1`
`y=-3(1/9)+3/3+1`
`y=-1/3+1+1`
`y=5/3`

Vertex `((-1)/3, 5/3)`

Find Intercepts

Y- intercept at `x=0`

`y=-3(0)-3(0)+1`

`(1, 0)`

X Intercept, at `(y=0)`

`-3x^2-3x+1=0`

`x=((-b)+-sqrt((-b)^2-(4ac)))/(2a)`
`x=((-(-3))+-sqrt((-3)^2-(4 xx(-3)xx1)))/(2 xx(-3))`
`x=(3+-sqrt(9-(-12)))/(-6)`
`x=(3+-sqrt(21))/(-6)`
`((3+sqrt(21))/(-6), 0)`
`((3-sqrt(21))/(-6), 0)`