How do you use the important points to sketch the graph of #y=2x^2+2#?

Question

950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-19T06:27:26

Please see the explanation below

The equation is

#y=2x^2+2#

Compare this equation to

#y=ax^2+bx+c#

The coefficient of #x^2# is

#a=2 >0#

So the curve is convex.

Now, calculate the intercepts with the y-axis

#x=0#, #=>#, #y=2#

So a point is #(0,2)#

Now, calculate the intercepts with the x-axis

#y=0#, #=>#, #2x^2+2=0#

#=>#, #x^2+1=0#

#x^2=-1#

There are no intercepts with the x-axis

When #x=-x#, the curve does not change

The axis of symmetry is #x=0#

And the minimum point is #(0,2)#

Now, you can sketch the curve

graph{2x^2+2 [-16.02, 16.01, -8.01, 8.01]}