Help with algebra II?

[Darth Cupcake]

I need to graph this equation:

y = |3x - 2| - 8

I know that it will be v-shaped.

The vertex she gave is (2/3,8)

My teacher wants the increase in the domain to be in thirds.

I want to plot the point 1/3 on my graph. How do I do this? I don't know she got the vertex from this. How do I plug in the x value of 1/3? I'm totally lost..

Thanks in advance!

DC

[tenor_singer]

Here's a good general equation to use when graphing absolute value:

f(x) = a|x - h| + k

If a>1, you will get a narrower "V" shape than you'd normally get if you graphed f(x) = |x|
If 0<a<1, you will get a wider "V" shape than you'd normally get if you graphed f(x) = |x|
If a < 0, you will get an inverted "V" shape than you'd normally get if you graphed f(x) = |x|
The vertex (point of the "V") is at point (h, k)

example:

f(x) = 3|x - 5| - 4

You'd get a skinny "V" shape whose point is located at (5, -4)


Now... if you are interested in specific points along the absolute value curve, all you need to do is substitute them into the function.

example:

Taking the previous example's function... if I want to find the point with a domain value of 3, all I'd do is find f(3).

f(3) = 3|[3] - 5| - 4
f(3) = 3|-2| - 4
f(3) = 3(2) - 4
f(3) = 6 - 4
f(3) = 2

So the point (3, f(3) ) or (3, 2) would lie on the "V".

I hope this helps!

Good luck!

PS... if there is a number in front of the "x" within the absolute value, all you have to do is factor it out to make the given function fit the general equation.

example:

f(x) = |3x + 8| + 5
f(x) = |3||x + (8/3) | + 5 <-- Factor out the 3.
f(x) = 3|x + (8/3) | + 5

So this would be a skinny "V" at point (-8/3, 5)

Again... good luck!