Question

How do you find the vertex, x and y intercepts for `f(x) = -3x^2 + 6x -2`?

Answer 1

Vertex`->(x.y)->(+1,?)`
Having shown you a cool trick to get the value of `x` I will let you finish it to find `y`

Explanation:

`color(blue)("To find "x_("vertex"))`

Given: `color(white)("xx")y=-3x^2+6x-2color(white)(.)` .......(1)

Write as: `-3(x^2color(brown)(-2)x)-2`

Now consider the `color(brown)(-2)" from the "-2x" inside the brackets"`

Multiply this value by `(-1/2)`

`(-1/2) xx (-2)=+1`

`color(blue)(x_("vertex")=+1)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare this to the graph:

Ok! You have now found the value for `x=1`

Substitute this back into equation (1) to find the value for `y`
I will let you do that!

Answer 2

Vertex (1, 1)
y = -2
`x = (3 +- sqrt3)/3`

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = -6/-6 = 1`
y-coordinate of vertex:
`f(1) = -3 + 6 - 2 = 1.`
Vertex (1, 1)
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make y = 0 and solve:
`f(x) = -3x^2 + 6x - 2 = 0`
`D = d^2 = b^2 - 4ac = 36 - 24 = 12` --> `d = +- 2sqrt3`
`x = -b/(2a) +- d/(2a) = 1 +- sqrt3/3 = (3 +- sqrt3)/3`