Question

How do you find the axis of symmetry and vertex point of the function: `Y=3x^2 + 8x + 4`?

Answer 1

The axis of symmetry is `-4/3`.
The vertex is `(-4/3,-4/3)`.

Explanation:

`y=3x^2+8x+4` is a quadratic equation in the form `ax^2+bx+c`, where `a=3, b=8, c=4`.

Axis of Symmetry
An imaginary vertical line that divides the parabola into to equal halves.

Equation: `x=(-b)/(2a)`

`x=(-8)/(2*3)=-8/6=-4/3`

Axis of symmetry is `x=-4/3`.

Vertex
The maximum or minimum point of the parabola. Since `a` is a positive number, the parabola opens upward, so the vertex is the minimum point.

The `x` value of the point is the axis of symmetry, `x=-4/3`. To find the `y` value, substitute `-4/3` for `x` in the equation and solve for `y`.

`y=3x^2+8x+4=`

`y=3(-4/3)^2+8(-4/3)+4=`

`y=3(16/9)-32/3+4=`

`y=48/9-32/3+4=`

The LCD is `9`. Make each term have a denominator of `9`.

`y=48/9-32/3*3/3+4*9/9=`

`y=48/9-96/9+36/9`

Add the numerators.

`y=(48-96+36)/9=`

`y=-12/9`

Simplify.

`y=-4/3`

The vertex is `(-4/3,-4/3)`

graph{y=3x^2+8x+4 [-10, 10, -5, 5]}