How do you find the vertex and intercepts for # y=(x+8)^2-2#?

1 Answer
Nov 17, 2015

All you do is look at the equation.


This is an equation for a parabola. We know this because the mother function (x^2) is in it. However, we can see that the original parabola has been undergone a few transformations. There is an easy way to graph this parabola. Let's start with the mother function.
x^2 on a graph is a parabola passing through the points (1,1) and (-1,1) with a vertex at (0,0).

Now, we first look at the (x+8)^2 term. Since the 8 is a number inside of the paranthesis, we know that it is a transformation on the x axis. We take the negative of 8 and voila, the parabola is moved 8 to the left (towards the negative side) of the graph. Your parabola should now have a vertex of (-8,0) and pass through the points (-7,1) and (-9,1).

Now we look at the -2 term. Since it is outside of the parenthesis, we know it is a shift on the y axis. Unlike an transformation on x, the y shift stays the same so since it is a -2 the parabola will be transferred down 2 on the y axis.

Your parabola should now have been moved to the left 8, and down two, producing a vertex of (-8,-2) and passing through the points (-7,-2) and (-9,-1). To find the x intercept, set your equation equal to 0 and solve for x.

It should look like this:
You would then take the square root of both sides and solve down for x, but I assume you know how to do that.

To find the y intercept, simply plug 0 into x and solve.

Alternatively, there is a much easier way of doing this. By looking at the equation, we know that the vertex is at (negative value of the number inside the parenthesis, value of the number outside the parenthesis) or as we already stated (-8,-2).

I hope this helps!