How do you graph #Y=2(x-1)^2+4#?

1 Answer
Aug 1, 2015

You need three points and then trace the curve.


this equation is in vertex form. Which means the most important point does not need to be calculated. The vertex form in general is:


where (h,k) is the vertex of the parabola.
For your equation that means the vertex is (1,4). Plot this point.

NOTE: there is a sign change for the h but not k.

Now we need another point. The easiest to find is by letting x=0

So the next point is (0,6)

Now we need one last point to sketch the curve. I always try to use the fact that the parabola is symmetric to find the last point. Notice that the new point is one to the left and two up.

#(1,4) -> (1-1, 4+2) -> (0,6)#

so the symmetric point is one to the right and two up.

#(1,4) -> (1+1, 4+2) -> (2,6)#

so our three points are (0,6) , (1,4) , (2,6). Plot them all and connect them with a smooth curve.

NOTE: you may need to pick one more x, further from the vertex, to make a smooth curve. so just follow what I did above for another x-value.

and that is how you graph a parabola in vertex form, and that makes us happy :)