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## Algebra 1

### Course: Algebra 1>Unit 10

Lesson 1: Graphs of absolute value functions

# Absolute value graphs review

The general form of an absolute value function is f(x)=a|x-h|+k. From this form, we can draw graphs. This article reviews how to draw the graphs of absolute value functions.
General form of an absolute value equation:
f, left parenthesis, x, right parenthesis, equals, start color #e07d10, a, end color #e07d10, vertical bar, x, minus, start color #11accd, h, end color #11accd, vertical bar, plus, start color #11accd, k, end color #11accd
The variable start color #e07d10, a, end color #e07d10 tells us how far the graph stretches vertically, and whether the graph opens up or down. The variables start color #11accd, h, end color #11accd and start color #11accd, k, end color #11accd tell us how far the graph shifts horizontally and vertically.
Some examples:

### Example problem 1

f, left parenthesis, x, right parenthesis, equals, vertical bar, x, minus, 1, vertical bar, plus, 5
First, let's compare with the general form:
f, left parenthesis, x, right parenthesis, equals, start color #e07d10, a, end color #e07d10, vertical bar, x, minus, start color #11accd, h, end color #11accd, vertical bar, plus, start color #11accd, k, end color #11accd
The value of start color #e07d10, a, end color #e07d10 is 1, so the graph opens upwards with a slope of 1 (to the right of the vertex).
The value of start color #11accd, h, end color #11accd is 1 and the value of start color #11accd, k, end color #11accd is 5, so the vertex of the graph is shifted 1 to the right and 5 up from the origin.
Finally here's the graph of y, equals, f, left parenthesis, x, right parenthesis:

### Example problem 2

f, left parenthesis, x, right parenthesis, equals, minus, 2, vertical bar, x, vertical bar, plus, 4
First, let's compare with the general form:
f, left parenthesis, x, right parenthesis, equals, start color #e07d10, a, end color #e07d10, vertical bar, x, minus, start color #11accd, h, end color #11accd, vertical bar, plus, start color #11accd, k, end color #11accd
The value of start color #e07d10, a, end color #e07d10 is minus, 2, so the graph opens downwards with a slope of minus, 2 (to the right of the vertex).
The value of start color #11accd, h, end color #11accd is 0 and the value of start color #11accd, k, end color #11accd is 4, so the vertex of the graph is shifted 4 up from the origin.
Finally here's the graph of y, equals, f, left parenthesis, x, right parenthesis:
Want more practice? Check out this exercise.

## Want to join the conversation?

• is there any easier steps to explain this type of lesson •   Maybe I can better explain

when you have an absolute value function you want to look at what are in the places of a, h and k. a|x-h|+k Specifically you want to look at h and k first. Normally the tip of the V shape is at (0,0) this changes depending on h and k. specifically it moves the tip to (h,k) so if you have |x+5|-7 then the tip of the V shape goes to (-5,-7). if you wonder why it is -5 even though we are adding 5, you just need to look at the original a|x-h|+k if we had -5 then it would be just like that, but since it is +5, we have to look at it as - -5, minus negative 5. so if it helps, the x coordinate is kinda backwards.

After the V tip you then look at a. treat it like a linear equation where a is the slope. so if a was -3 that's down 3 right 1 using rise over run. then, since it's an absolute value function you need to know that the same line goesalong the left to make that V shape, so -5 would mean on the left down 3 and left 1.

if you ever have something like a|bx-h|+k where there is a number in front of the x you need to get rid of it if you are not aware of factoring this is what it would look like a|b||x - h/b|+k where a|b| becomes the new "a" and h/b becomes the new h, then you would solve it normally. The point being you always want x by itself for this. Also, keep in mind that even if inside the absolute value bars if b was negative, outside it becomes positive.

Let me know if that didn't help, or if there is a specific function you are struggling with, or maybe would even like some to try out.
• In example problem 1, why isn’t the graph shifted 1 unit to the left instead of to the right? • It is shifted to the right because x-1 would make it 0 when x=1 because
x=1
1-1=0
So, we always want the absolute value part of the equation to be equal to 0 when we use x as the horizontal shifting.

While, the vertical part goes up with + not down because when,
y=a∣x−h∣+k
y-k=a|x-h|
So basically we transpose it to make it easier to distinguish.
• If someone needs:
Horizontal shift : y = f(x+b)
Vertical shift: y = f(x) +d
Reflection about the X-axis : y = -f(x)
Reflection about the Y-axis : y = f(-x)
Stretch/Compress in the X direction: y = f(a * x)
Stretch/Compress in the Y direction: y=f(x) * a • How would we utilize this in real life? For what careers? • How do you identify the vertex y intercept and x intercept • Hey there,

I'm not an expert here, but it was an interesting exercise to figure out the answer to your questions and I figured I might as well post it here. Sorry if it's too much of a wall of text to get through.

Just to recapitulate, the general form is:
f(x) = a|x−h| + k

The vertex is located at point (h,k). The minimum or maximum (depending on whether a is positive or negative) of the graph is at the point where x - h = 0. This is the same as saying x = h, which gives us the x-coordinate of the vertex. As for the y-coordinate: since we just saw that |x-h| = 0, a|x−h| must also be 0, which only leaves us with k.

To find the y-intercept, we can set x to 0. In the general formula, that means:

f(0) = a|0−h| + k
f(0) = a|h| + k

Which gives us, as a general rule, (0,(a|h|+k)) as the y-intercept. Taking one of the examples, f(x)= |x−1| + 5 where a=1, h=1 and k=5: the y-coordinate of the intercept is 1|1| + 5 = 6, which means the intercept is at (0,6).

The method I thought of to find the x-intercepts a bit more involved, maybe someone else knows an easier way. I basically just used algebra.

There can be 0 or 2 x-intercepts depending on the value of k and a.
There will be two x-intercepts if:
k > 0 and a < 0
or
k < 0 and a > 0
and no x-intercepts otherwise. That said, let's use the general form again and set the result of the function to 0 and try to solve for x.

0 = a|x−h| + k
-k = a|x−h|
-k/a = |x-h|
|-k/a| = x-h
|-k/a| + h = x
Here it gets a bit tricky. There can be two possible values such that their absolute value together with h adds up to x: -k/a and k/a, since both evaluate to the same absolute value. But since we're looking for two intercepts, it actually makes sense that there are two possible results for x:
-k/a + h = x and k/a + h = x

I tested this with f(x)= -2|x+5| +4
According to my result, -(4/-2) +(-5) and 4/-2 +(-5) should be the x-coordinates of this graph's y-intersects: -3 and -7, and it checks out! It kind of makes sense as well: we're dividing k - the difference in y from the x-axis at the maximum - by the slope. This should give as result the difference in x from the maximum, and then we're adding the amount by which the maximum was shifted.

Cheers if someone actually read all of these words.
• So is h is positive that means that it is actually negative? Is that why if its x + 3 on the graph you go to negative 3? • No, if it is positive it means I move in the negative direction, but if h is negative I move in the positive direction, it does not change the sign of h. The idea is that what value of x would make the inside of the absolute value (or other function) 0, so if you have x + 3, it would require x = -3 to be 0, thus causing a shift in the negative direction. The other idea is that since the formula has | x - h | + k where (h,k) is the vertex, then using x+ 3 would actually be x - (- 3) so -3 would cause a shift to the left.
• I am confused on how you know if the vertex is a minimum or maximum point.
(1 vote) • i dont know what made the diffrence to make it go up or down
(1 vote) • What if there's something in front of the x? In my homework, there's a problem that says a(x) = |5x|. How would I graph that?  