Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `g(x) = 3x^2 + 12x + 15`?

Answer 1

Vertex: `(-2,3)`
Axis of Symmetry: `x=-2`
Domain: `RR`
Range: `RR >=3`

Explanation:

The "vertex form" for a parabola is
`color(white)("XXX")g(x)=m(x-a)^2+b`
`color(white)("XXXXXX")`with the vertex at `(a,b)`

To convert `g(x)=3x^2+12x+15` into "vertex form"

Extract the `m`
`color(white)("XXX")g(x)=3(x^2+4x)+15`
Complete the square
`color(white)("XXX")g(x)=3(x^2+4x+2^2)+15 -3(2^2)`
`color(white)("XXX")g(x)=3(x+2)^2+3`
In complete vertex form:
`color(white)("XXX")g(x)=3(x-(-2))^2+3`
with the vertex at `(-2,3)`

The equation is of the form for a parabola with a vertical axis of symmetry opening upward and since the axis of symmetry passes through the vertex,
the axis of symmetry is `x=-2`

`g(x)=3x^2+12x+15` is defined for all Real values of `x`,
therefore the Domain is `x in RR`.

`g(x)` has values for all points at or above the vertex,
therefore the Range is `g(x) in RR >=3`
graph{3x^2+12x+15 [-7.85, 6.196, -0.84, 6.186]}