Question

How do you find the vertex and the intercepts for `f(x)=1-12x-3x^2`?

Answer 1

`y_("intercept")=+1`
`x_("intercepts") = +0.0817" and " -4.0827` to 4 decimal places

Vertex`->(x,y)-> (-2,+13)`

Explanation:

I much prefer the format:

`f(x)=-3x^3-12x+1`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Straight off, the "y_("intercept")" is the constant of +1")`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Write as `y=-3(x^3+12/3 x) +1`

`color(blue)(x_("vertex") = (-1/2)xx(+12/3) = -2)`

`color(brown)("By substitution of "x)`
`color(blue)(y_("vertex") =-3(-2)^2-12(-2)+1 = +13)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine x intercepts")`
This equation does not have whole number x intercepts

Set `0=-3x^2-12x+1`

Std form `-> y=ax^2+bx+c`

Where`" "x=(-b+-sqrt(b^2-4ac))/(2a)`

`=> x= (+12+-sqrt((-12)^2-4(-3)(1)))/(2(-3))`

`x= (12+-sqrt(156))/(-6)`

`x= -2+-sqrt(2^2xx39)/(-6)`

`x= -2+-sqrt(39)/(-3)`
'~~~~~~~~~~~~~~~~~~~~~~~
Consider: `color(blue)(x= -2-sqrt(39)/(-3) = +0.0817)` to 4 decimal places

Consider: `color(blue)(x= -2+sqrt(39)/(-3) = -4.0827)` to 4 decimal places