Question

How do you find the axis of symmetry, graph and find the maximum or minimum value of the function `f(x)= 3x^2`?

Answer 1

Refer to the explanation.

Explanation:

`f(x)=3x^2` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=3`, `b=0`, and `c=0`

Axis of symmetry: the vertical line that divides the parabola into two equal halves. For a quadratic equation in standard form, the formula for the axis of symmetry is:

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

Plug in the known values.

`x=(0)/(2*3)`

Simplify.

Axis of symmetry: `x=0`

This means that the axis of symmetry is on the x-axis, where `x=0`. This is also the `x`-value of the vertex.

Vertex: the maximum or minimum point on the parabola.

To determine the `y`-value of the vertex, substitute `0` for `x` and solve for `y`.

`y=3x^2`

Plug in `0` for `x`.

`y=3(0)^2=0`

Vertex: `(0,0)`

Since the vertex does not cross the x-axis, we don't have x-intercepts in terms of `(x,0)`, but we can determine other points of `(x,y)` on the parabola.

I will propose six values for `x` (3 positive and 3 negative), and plug them into the equation and solve for `y`. This will give six symmetrical points on the parabola.

`x=0.5`

`y=3(0.5)^2=0.75`

Point 1: `(0.5,0.75)`

`x=-0.5`

`y=3(-0.5)^2=0.75`

Point 2: `(-0.5,0.75)`

`x=1`

`y=3(1)^2=3`

Point 3: `(1,3)`

`x=-1`

`y=3(-1)^2`

`y=3`

Point 4: `(-1,3)`

`x=2`

`y=3(2)^2`

`y=12`

Point 5: `(2,12)`

`x=-2`

`y=3(-2)^2`

`y=12`

Point 6: `(-2,12)`

Plot the vertex and the other six points. Sketch a parabola through the points. Do not connect the dots.

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