Question

What is the the vertex of `y =3x^2+4x-18`?

Answer 1

`x_("vertex")=-2/3" "`I will let the reader find `""y_("vertex")`

Explanation:

Given:`" "y=3x^2+4x-18" "`..................................(1)

Write as:`" "y=3(x^2+4/3x)-18`

Using the `+4/3" from "(x^2+4/3x)`

`(-1/2)xx4/3 =-4/6=-2/3`

`color(blue)(x_("vertex") = -2/3)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`-2/3" " =" " -0.6666bar6" "=" "-0.6667` to 4 decimal places

`color(brown)("All you have to do now is substitute "x=-2/3" into")` `color(brown)("equation (1) to find "y_("vertex"))`

Tony B

Answer 2

May be done as follows

Explanation:

The given equation is
`y=3x^2+4x-18`
`=>y=3[x^2+2x(2/3)+(2/3)^2-(2/3)^2 -6]`

`=>y=3[(x+2/3)^2-(2/3)^2 -6]`
`=>y=3[(x+2/3)^2-4/9- 6]`
`=>y=3[(x+2/3)^2-58/9 ]`
`=>y=3(x+2/3)^2-58/9*3`
`=>y+58/3=3(x+2/3)^2`
putting ,`y+58/3=Y and x+2/3=X` we have
new equation
`Y =3X^2`,which has coordinate of vertex (0,0)
So putting X=0 and Y=0 in the above relation
we get
`x=-2/3`
and `y=-58/3 =-19 1/3`

so the actual coordinate of vertex is `(-2/3,-19 1/3)`