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

Dec 28, 2017

Refer to the explanation.

#### Explanation:

$f \left(x\right) = 3 {x}^{2}$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

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:

$x = \frac{- b}{2 a}$

Plug in the known values.

$x = \frac{0}{2 \cdot 3}$

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 $y$.

$y = 3 {x}^{2}$

Plug in $0$ for $x$.

$y = 3 {\left(0\right)}^{2} = 0$

Vertex: $\left(0 , 0\right)$

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

$x = 0.5$

$y = 3 {\left(0.5\right)}^{2} = 0.75$

Point 1: $\left(0.5 , 0.75\right)$

$x = - 0.5$

$y = 3 {\left(- 0.5\right)}^{2} = 0.75$

Point 2: $\left(- 0.5 , 0.75\right)$

$x = 1$

$y = 3 {\left(1\right)}^{2} = 3$

Point 3: $\left(1 , 3\right)$

$x = - 1$

$y = 3 {\left(- 1\right)}^{2}$

$y = 3$

Point 4: $\left(- 1 , 3\right)$

$x = 2$

$y = 3 {\left(2\right)}^{2}$

$y = 12$

Point 5: $\left(2 , 12\right)$

$x = - 2$

$y = 3 {\left(- 2\right)}^{2}$

$y = 12$

Point 6: $\left(- 2 , 12\right)$

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]}