Main content

## Algebra 1

### Course: Algebra 1 > Unit 10

Lesson 1: Graphs of absolute value functions- Shifting absolute value graphs
- Shift absolute value graphs
- Scaling & reflecting absolute value functions: equation
- Scaling & reflecting absolute value functions: graph
- Scale & reflect absolute value graphs
- Graphing absolute value functions
- Graph absolute value functions
- Absolute value graphs review

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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:

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

We're asked to graph:

First, let's compare with the general form:

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

We're asked to graph:

First, let's compare with the general form:

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 to learn more about absolute value graphs? Check out this video.*

*Want more practice? Check out this exercise.*

## Want to join the conversation?

- is there any easier steps to explain this type of lesson(9 votes)
- 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.(60 votes)

- In example problem 1, why isn’t the graph shifted 1 unit to the left instead of to the right?(8 votes)
- 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.(18 votes)

- 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(13 votes) - How would we utilize this in real life? For what careers?(4 votes)
- mathematician(11 votes)

- How do you identify the vertex y intercept and x intercept(2 votes)
- 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.(9 votes)

- 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?(3 votes)
- 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.(5 votes)

- I am confused on how you know if the vertex is a minimum or maximum point.(1 vote)
- The general form of the absolute value function is:

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

When "a" is negative, the V-shape graph opens downward and the vertex is the maximum.

When "a" is positive, the V-shape graph opens upward and the vertex is a minimum.

Hope this helps.(7 votes)

- i dont know what made the diffrence to make it go up or down(1 vote)
- y=a|x-h|+k

here 'a' is the slope of the line. If slope (which is 'a') is positive the line will go upward. As we are dealing with the absolute of x (|x|) it won't affect the value of 'a'.(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?(2 votes)
- You would simplify the expression inside the
`| |`

and take the absolute value of the result. For example

|5*0| = |0| = 0

|5*1| = |5| = 5

|5*-1| = |-5| = 5

|5*2| = |10| = 10

|5*-2| = |-10| = 10(2 votes)

- what if x has a coefficient(1 vote)
- Great Question. When x has a non-one and non-zero coefficient, the curve stretches or shrinks.

When coefficient of x is larger than one, then the curve shrinks along the x-axis with the scale of 1 / (coefficient).

When coefficient of x is smaller than one but larger than zero, the curve expands with the scale of 1 / (coefficient).

If it's negative then it's a reflection.(5 votes)