How do you graph $y = 2{x^2} + 4$?

Hint:According to the given question, we have to draw the graph $y = 2{x^2} + 4$.
So, first of all we have to compare the given parabolic equation as $y = 2{x^2} + 4$ with the standard form of parabolic equation as $a{x^2} + bx + c$and find the value of $a$, if $a$ is positive then the parabola will be open ups and if $a$ is negative then the parabola will be open down.
Now, we have to write this parabolic equation in the standard form of the parabolic equation as $a{x^2} + bx + c$.
Now, we have to find the vertex of the given parabolic equation as $y = 2{x^2} + 4$ by finding the x-coordinates of the vertex with the help of the formula as mentioned below.
Formula used:
X-coordinates of vertex$ = \dfrac{{ - b}}{{2a}}$, where $a$ is the coefficient of ${x^2}$ and $b$ is the coefficient of $x$.
Now, we have to find the y-coordinates of the vertex by putting $x = 0$ in the given parabolic equation.
Now, we have to find the different y-intercept on the given parabola by putting the value of $x$ as 1,-1, 3,-3 and plot the graph.

Complete step by step answer:
Step 1: First of all, we have to write this parabolic equation in the standard form of the parabolic equation as $a{x^2} + bx + c$.
$ \Rightarrow y = 2{x^2} + 0x + 4$
Step 2: Now, we have to compare the given parabolic equation as obtained in the solution step 1 with the standard form of the parabolic equation as $a{x^2} + bx + c$ and find the value of$a$.
So, here we can see that the value of $a$$\left( 2 \right)$ is positive so parabola will be open ups.
Step 3: Now, we have to find the vertex of the given parabolic equation as $y = 2{x^2} + 4$ by finding the x-coordinates of vertex with the help of the formula as mentioned in the solution hint.
X-coordinates of vertex$ = \dfrac{0}{{2 \times 2}} = 0$
Now, we have to find the y-coordinates of the vertex by putting $x = 0$ in the given parabolic equation as $y = 2{x^2} + 4$.
Y-coordinates of the vertex$ = 2 \times {\left( 0 \right)^2} + 4 = 4$
So, the vertex of the given parabola is $\left( {0,4} \right)$
Step 4: we have to find the different y-intercept on the given parabola by putting the value of $x$ as 1,-1, 3,-3 and plot the graph.
So, at $x = 1$, Y-intercept$ = 2 \times {\left( 1 \right)^2} + 4$
$ \Rightarrow 6$
So, at$x = - 1$, Y-intercept$ = 2 \times {\left( { - 1} \right)^2} + 4$
$ \Rightarrow 6$
So, at$x = 3$, Y-intercept$ = 2 \times {\left( 3 \right)^2} + 4$
$ \Rightarrow 22$
So, at $x = - 3$, Y-intercept$ = 2 \times {\left( { - 3} \right)^2} + 4$
$ \Rightarrow 6$
Step 5: Now, we have find the different points on the parabola from the solution step 4, as $\left( {1,6} \right),\left( { - 1,6} \right),\left( {3,22} \right),\left( { - 3,22} \right)$

Final solution: Hence, the graph of the given parabola as $y = 2{x^2} + 4$,


Note:
-It is necessary to check if the given parabola in the question will be open ups or open down by comparing the given parabola with the standard form of the parabola.
-It is necessary to find the different points on the parabola by putting $x = 1, - 1,3, - 3$ in the given parabolic equation.