Question

How do you find the intercepts, vertex and graph `f(x)=-3x^2-4x`?

Answer 1

See graph.

Explanation:

Set `y=0` solve for `x` by factoring to find the x-intercept(s) if they exist (remember `f(x) = y`):

`f(x)=-3x^2-4x`

`0=-3x^2-4x`

`0=x(-3x-4)`

`x=0` and `x=-4/3`

Set `x=0` solve for `y` to find the y-intercept:

`f(x)=-3*0^2-4*0 = 0`

`y = 0`

in the form:

`ax^2 + bx +c` the formula for axis of symmetry is:

`aos = (-b)/(2a)`

`f(x)=-3x^2-4x +0`

a=-3
b=-4

`aos = (-(-4))/(2*-3)`

`aos = -2/3`

The formula for the vertex is:

vertex `(x,y) = (aso, f(aos))`

`f(-2/3)=-3(-2/3)^2-4*(-2/3)+0`

`f(-2/3)=4/3`

vertex = #(-2/3, 4/3)

Finally since `a<0` the Parabola opens down and has a maximum at the vertex:

graph{-3x^2-4x [-5, 5, -2.5, 2.5]}