Question

How do you find the vertex and intercepts for `f(x)=3x²+12x+4`?

Answer 1

`color(blue)("Vertex "->(x,y)->(-2,-8)`
`color(blue)(y_("intercept")" at "x=0 -> y=4)`

`color(blue)(x_("intercepts"):`
`color(green)(x=+-sqrt(24)/3-2" Exact values")`

`color(green)(x~~3.633" and "-0.367" Approx values")`

Explanation:

Given;`" "y=3x^2+12x+4`

`color(blue)(y_("intercept")" at "x=0 -> y=4)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine "y_("intercept") " & Vertex"`

Transpose equation into vertex form
Standard form `y=ax^2+bx+c`
Vertex form `y=a(x+b/(2a))^2+c - [(b/2)^2]`

`y=3(x+2)^2+4 -12`

`y=3(x+2)^2-8`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(x_("vertex")->(-1)xxb/(2a)" "->" "(-1)xx(2) = -2)`

`color(blue)(y_("vertex")-> c - [(b/2)^2]" "->" "-8`

`color(blue)("Vertex "->(x,y)->(-2,-8)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine " x_("intercepts"))`

Set `" "color(brown)(y=3(x+2)^2-8)" to "color(green)( 3(x+2)^2-8 =0)`

`(x+2)^2=8/3`

Take square roots of both sides

`x+2=+-sqrt(8/3)`

`x= +-sqrt(8/3)-2" " -> sqrt(8)/sqrt(3)xxsqrt(3)/sqrt(3) = sqrt(24)/3`

`x=+-sqrt(24)/3-2" "` Exact values

`x~~3.633" and "-0.367`