91Ó°ÊÓ

We have $f\left(x\right)=5x^2+2$.

The parabola will open upwards here, because the leading coefficient is positive. Then finding the axis of symmetry, vertex, the intercepts, will help us to plot the points of the graph and get the required graph.

Axis of symmetry is a vertical line given by $x=\frac{-b}{2a}$where $a,b,c$are the coefficients of the equation when compared with the standard form of the quadratic equation, i.e., $a{x}^2+bx+c$.

So the axis of symmetry comes out to be,
$$\begin{gathered} x=\frac{-b}{2a} \\ x=-\frac{0}{2\left(5\right)} \\ x=0 \end{gathered}$$

Thus the equation of axis of symmetry is $x=0$.

Vertex is a point on the line of axis of symmetry. Here axis of symmetry is $x=0$.

We will get the vertex by putting $x=0$in the equation. This gives us,

$$\begin{gathered} =5x^2+2 \\ =5{\left(0\right)}^2+2 \\ =0+2 \\ =2 \end{gathered}$$

We get the vertex as$(0,2)$.

To find $y-$intercept, we need to keep $x=0$in the equation. This gives us,

$$\begin{gathered} =5x^2+2 \\ =5{\left(0\right)}^2+2 \\ =0+2 \\ =2 \end{gathered}$$
The intercept comes out to be$(0,2)$.

To find $x-$intercept, we need to keep $f\left(x\right)=0$and solve for $x$.

$$\begin{gathered} 5x^2+2=0 \\ 5x^2=-2 \\ {x}^2=\frac{-2}{5} \end{gathered}$$
As we know that square of a number can never be negative. So we can conclude that this function has no$x-$intercept.

Plotting all the points we got and the axis of symmetry give us a graph. Joining these points will give us the graph as-