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

Aug 1, 2015

You need three points and then trace the curve.

#### Explanation:

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

$y = a {\left(x - h\right)}^{2} + k$

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

$y = 2 {\left(0 - 1\right)}^{2} + 4$
$y = 2 {\left(- 1\right)}^{2} + 4$
$y = 2 \left(1\right) + 4$
$y = 2 + 4$
$y = 6$
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.

$\left(1 , 4\right) \to \left(1 - 1 , 4 + 2\right) \to \left(0 , 6\right)$

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

$\left(1 , 4\right) \to \left(1 + 1 , 4 + 2\right) \to \left(2 , 6\right)$

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 :)