How do you find the important points to the graph $f(x) = 2{x^2}$?

Hint: The following question asks us to plot the given curve and in the process find the important points of the graph represented by the above function. The above function is the formula of a parabola since the coefficient of ${x^2}$ here is positive the parabola will be upward shaped and there is no coefficient of ${y^{\tilde 2}}$. Then we will find the vertex and the nature of the graph and also plot it.

Complete step by step solution:
The given equation is an equation of the parabola since the coefficient of ${x^2}$ is positive the graph will be upward facing. Then we will find the $y$ intercept:
For $y$ intercept putting $x = 0$ we get,
$y = 0$ so the y intercept is zero
And also $x$ intercept is zero we can tell that by putting the value of $y = 0$ in the above function.
$0 = 2{x^2}$
Gives $x = 0$ therefore the parabola passes through origin. The given equation thus is of a normal parabola with following attributes :
It is upward facing and symmetric along the axis of $y$.
The given conic also passes through the origin thus the important point of the given graph can be said to be as the origin from which it passes and touches both the axis at that point . We will also plot the given conic onto a graph the graph is given below:

Note: If the sign of the coefficient of ${x^2}$ would have been negative then the given parabola would have been downward facing all the other important attributes would have been same.