How do you sketch the graph of #y=(x+3)^2+6# and describe the transformation?

1 Answer
Shantelle
Dec 17, 2017

Refer to the explanation.

Explanation:

First, let's look at the transformations of this equation.

When something is with the #x#, in this case #x+3#, then you will do the opposite of what it says, because when you set #x + 3# to equal to #0#, then you will get #-3#, which is the opposite of #3#.

However, when something is "outside" the #x#, in this case #6#, then that applies to the y-values and it just does what it says. So the y-values would always add #6#.

Here is the graph (should be arrows at each end, just doesn't show):
enter image source here

As you can see, there is a point, or the vertex, is at #(-3, 5)# as our transformation showed. Then we form a parabola.

When the coefficient of a quadratic equation is positive , (ex: #x^2#, #10x^2#) then the parabola will face up .

When the coefficient is negative, (ex: #-x^2#, #-10x^2#), then the parabola will face down .

Our coefficient is just #x#, so that is positive, therefore the graph grows up.

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