Question

How do you graph the parabola \[y=2{{x}^{2}}+4\] using vertex, intercepts and additional points.

Answer 1

Hint: Now first we will compare the given equation with $y=a{{\left( x-h \right)}^{2}}+k$ and find the values of a, h and k. Now we have if a is positive then the parabola is upward facing and similarly if a is negative then the parabola is downward facing. Also $\left( h,k \right)$ is the vertex of the parabola. Now we will find some points which lie on the curve and plot them on a graph and draw a parabola.

Complete step-by-step solution:
Parabola is curve represented by an equation in the form $a{{x}^{2}}+bx+c$
Now to plot the parabola we must find the vertex of the parabola and some points which lie on the parabola.
The more points we get on the curve the more precise the graph will be.
Now the equation of parabola can be written in the form of $y=a{{\left( x-h \right)}^{2}}+k$ where $\left( h,k \right)$ is the vertex of the parabola.
Now let us compare the given equation \[y=2{{x}^{2}}+4\] to $y=a{{\left( x-h \right)}^{2}}+k$ we get h = 0 and k = 4.
Hence the vertex of the parabola is $\left( 0,4 \right)$
Now for the parabola $y=a{{\left( x-h \right)}^{2}}+k$ we have if a > 0 then the parabola is upwards facing and if a < 0 then the parabola is downwards facing.
Now from the given equation we can say that a = 2 > 0. Hence the parabola is upward facing parabola.
Now let us find some points on the parabola by.
Now if we substitute x = 0 in the equation we get, y = 4. This is nothing but an intercept of the equation.
Now let us substitute $x=1$in the equation. Then we get $y=2{{\left( 1 \right)}^{2}}+4=6$
Similarly by substituting $x=-1$in the equation we get, $y=2{{\left( -1 \right)}^{2}}+4=6$
Now again let us substitute $x=2$ we get, $y=2{{\left( 2 \right)}^{2}}+4=12$ and if we substitute $x=-2$ we get $y=2{{\left( -2 \right)}^{2}}+4=12$
Hence we get $\left( 0,4 \right),\left( 1,6 \right),\left( -1,6 \right),\left( 2,12 \right)$ and $\left( -2,12 \right)$ are the points which lie in the parabola with vertex $\left( 0,4 \right)$.
Now let us plot these points and draw a parabola.

Note: Note that here we will not have x intercept since the graph has vertex on y axis and the parabola is upward facing hence it will never touch x axis. Even if we substitute y = 0 we will get no real values for x. Also while plotting the points take negative and positive both values for the precise graph.