The absolute value parent function and its vertex, translating a graph horizontally and vertically, stretching, shrinking and reflecting it, the combined general form, and writing an equation from a graph.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Use Absolute Value Functions and Transformations
Objectives
Five outcomes. The third one has a sign convention that catches almost everybody once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-127 — the lesson these objectives are drawn from
Warm-up
Lesson 1.7 defined absolute value as distance. This lesson makes it a function and draws it.
Discussion prompt
Evaluate the absolute value of x at x equal to -3, -1, 0, 1 and 3. Plot those five points in your head. What shape do they make, and why is it symmetric?
Hint: Compare the value at 3 with the value at -3.
Answer:
\[ 3, \; 1, \; 0, \; 1, \; 3 \]
A V, with its point at the origin. It is symmetric because a number and its opposite are the same distance from zero, so the graph gives them the same height — which is exactly what the distance definition from Lesson 1.7 says.
Concept
Every absolute value function in this lesson is the same V, shifted, stretched, or flipped. Recognising a family and its parent — the habit begun with lines in Lesson 2.3 — means you never have to plot a table of values again.
transformation — A change to a graph's size, shape, position or orientation. A translation is the kind that shifts a graph without changing its size or shape.
\[ f(x) = \lvert x \rvert \]
The highest or lowest point of an absolute value graph is its vertex. For the parent it is the origin.
Figure (svg): The V-shaped graph of the absolute value parent function with its vertex at the origin and its two straight branches labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-123 — Parent Function for Absolute Value Functions
Section
Section 1
Concept
To the right of zero the graph is the line y equals x; to the left it is the line y equals negative x. The two meet at the origin, which is the vertex, and the graph is symmetric about the vertical axis.
vertex — The highest or lowest point on the graph of an absolute value function. For the parent function it is the origin.
\[ f(x) = \lvert x \rvert = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} \]
So an absolute value graph is not curved. It is two straight pieces, and everything you know about slopes applies to each piece separately.
Figure (svg): The V-shaped graph of the absolute value parent function with its vertex at the origin and its two straight branches labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-123
Picture it
The three-case definition from Lesson 1.7, drawn.
Figure (svg): The V-shaped graph of the absolute value parent function with its vertex at the origin and its two straight branches labelled
The right branch has slope 1 and the left branch slope negative 1. That pair of slopes is what the coefficient a will change in Section 3.
Worked example
Five points, and the shape follows.
\[ \text{Graph } f(x) = \lvert x \rvert \text{ from a table of five values.} \]
Evaluate at two negative inputs
Why: The third case applies: the absolute value of a negative is its opposite, so -3 gives 3 and -1 gives 1.
\[ (-3, 3), (-1, 1) \]
Evaluate at zero
Why: Zero units from zero.
\[ (0, 0) \]
Evaluate at two positive inputs
Why: The first case applies: a positive number is its own absolute value.
\[ (1, 1), (3, 3) \]
Plot and join
Why: The points lie on two straight half-lines meeting at the origin.
\[ a V\text{ with vertex } (0, 0) \]
Figure (svg): The solution to Worked example build the parent graph from the definition shown as a ladder of expressions, one row per algebraic move
\[ f(x) = \lvert x \rvert \]
Verify: check the symmetry claim on a pair
Why: The point (3,3) is on the graph, and so is (-3,3): same height, opposite inputs. That is what symmetry about the vertical axis means, and it holds for every pair because a number and its opposite are equally far from zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-123
Sorting
Test each point against y equals the absolute value of x.
Sort into buckets
Sort each point by whether it lies on the parent graph.
The whole graph lives at or above the horizontal axis, which is the graphical face of the fact that a distance cannot be negative.
Worked example
Applying Lesson 2.2 to each half separately.
\[ \text{Find the slope of each branch of } y = \lvert x \rvert. \]
Take two points on the right branch
Why: The points (1,1) and (3,3), both with positive inputs.
\[ (1, 1)\text{ and } (3, 3) \]
Compute that slope
Why: Three minus one over three minus one.
\[ m = \frac{2}{2} = 1 \]
Take two points on the left branch
Why: The points (-3,3) and (-1,1), both with negative inputs.
\[ (-3, 3)\text{ and } (-1, 1) \]
Compute that slope
Why: One minus three is negative two; negative one minus negative three is two.
\[ m = -\frac{2}{2} = -1 \]
Figure (svg): The solution to Worked example the slopes of the two branches shown as a ladder of expressions, one row per algebraic move
\[ \text{right: } m = 1 \qquad \text{left: } m = -1 \]
Verify: check against the piecewise definition
Why: For positive x the rule is y equals x, whose slope is 1; for negative x it is y equals negative x, whose slope is negative 1. The computed slopes match the definition exactly, which confirms the graph is two lines rather than a curve.
Trap
\[ f(x) = \lvert x \rvert \]
Draw a smooth U through the plotted points
Why: The vertex is treated as if it were the bottom of a parabola.
A smooth curve at the bottom would mean the slope changes gradually — but on this graph the slope is exactly negative one, then exactly one, with nothing in between.
\[ f(x) = \lvert x \rvert \]
Draw two straight segments meeting at a sharp corner
Why: Each branch is a line with a constant slope, so neither can bend.
\[ \text{left: slope } -1 \qquad \text{right: slope } 1 \]
The corner at the vertex is genuine. Compare y equals x squared, which really is a smooth U — a distinction Chapter 4 will need.
Explain it to yourself
Most graphs you have met so far are smooth or straight throughout.
\[ f(x) = \lvert x \rvert \]
Discussion prompt
Explain why this graph has a sharp corner at the origin rather than a smooth turn. What happens to the slope as you cross x equal to zero, and why does that force a corner?
Hint: Compute the slope just left of zero and just right of zero.
Answer:
Just left of zero the slope is exactly negative one, and just right of zero it is exactly one. There is no gradual change between them — the slope jumps.
A smooth turn would require the slope to pass through every value between negative one and one, which would take some horizontal distance. Here it happens at a single point, and a slope that jumps at a point is what a corner IS.
Matching
The V is two lines in disguise.
Match the pairs
Why: Two of these are the pieces of the piecewise definition from Lesson 1.7, now read as graphs. The symmetry rule says the graph is unchanged by reflecting it across the vertical axis, which is what makes the two branches mirror images with opposite slopes.
Prediction
Commit before reasoning.
Predict first
What is the range of the parent function f of x equals the absolute value of x?
Correct: All non-negative real numbers — zero and everything above it.
\[ \text{domain: all reals} \qquad \text{range: } y \geq 0 \]
Why: An absolute value is a distance, so it can never be negative, which rules out everything below the axis. But it CAN be zero, at x equal to zero, so the range includes zero rather than excluding it. The domain is a separate question and is all real numbers, since every number has a distance from zero.
Section
Section 2
Concept
The graph of y equals the absolute value of x minus h, plus k, is the parent V slid h units horizontally and k units vertically. Its vertex is at the point (h, k), and its shape is unchanged.
\[ y = \lvert x - h \rvert + k \;\Longrightarrow\; \text{vertex } (h, k) \]
The h is subtracted inside the bars, so an equation reading x plus 4 has h equal to negative 4 and shifts LEFT. The k is added outside and behaves as written.
Figure (svg): The parent V and a translated copy, with the horizontal and vertical shifts marked and the new vertex labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-124 — Translations
Picture it
Example 1: y equals the absolute value of x plus 4, minus 2.
Figure (svg): The parent V and a translated copy, with the horizontal and vertical shifts marked and the new vertex labelled
Rewriting it as x minus negative four makes h visible as negative four. The textbook suggests exactly that rewrite, and it is worth doing every time until the flip is automatic.
Worked example
Example 1, all four steps.
\[ \text{Graph } y = \lvert x + 4 \rvert - 2 \text{ and compare it with } y = \lvert x \rvert. \]
Rewrite it to expose h
Why: The form subtracts h, so x plus 4 is x minus negative four, making h equal to negative four and k equal to negative two.
\[ y = | x - (-4) | + (-2) \]
Plot the vertex at (h, k)
Why: Negative four, negative two.
\[ \text{vertex } (-4, -2) \]
Plot another point and use symmetry
Why: Going two right and two up from the vertex gives (-2, 0); the mirror image is (-6, 0).
\[ (-2, 0)\text{ and } (-6, 0) \]
Join with a V and compare
Why: The parent V slid left four and down two, with the same shape.
\[ \text{left } 4,\text{ down } 2 \]
Figure (svg): The solution to Worked example graph a translated V shown as a ladder of expressions, one row per algebraic move
\[ y = \lvert x + 4 \rvert - 2, \quad \text{vertex } (-4, -2) \]
Verify: substitute the vertex and one branch point
Why: At x equal to negative four the expression inside is zero, so y is negative two — the vertex. At x equal to negative two the inside is 2, so y is 2 minus 2, which is 0 — matching the plotted point. Both check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 124-124
Matching
Find the value of x that makes the inside zero.
Match the pairs
Why: In every case the first coordinate is the value of x that makes the expression inside the bars equal zero, and the second is the constant outside. The two with a plus sign inside have negative first coordinates, which is the flip — and asking what makes the inside zero produces it automatically without any rule to remember.
Worked example
Guided Practice 1. Both signs behave the other way this time.
\[ \text{Graph } y = \lvert x - 2 \rvert + 5 \text{ and compare it with } y = \lvert x \rvert. \]
Read h and k directly
Why: The form already subtracts inside and adds outside, so h is 2 and k is 5.
\[ h = 2, k = 5 \]
Plot the vertex
Why: At the point (2, 5).
\[ \text{vertex } (2, 5) \]
Plot symmetric points
Why: One right and one up gives (3, 6); the mirror is (1, 6).
\[ (3, 6)\text{ and } (1, 6) \]
Describe the translation
Why: The parent V slid right two and up five.
\[ \text{right } 2,\text{ up } 5 \]
Figure (svg): The solution to Worked example a shift right and up shown as a ladder of expressions, one row per algebraic move
\[ y = \lvert x - 2 \rvert + 5, \quad \text{vertex } (2, 5) \]
Verify: check that the graph never dips below its vertex
Why: The absolute value part is at least zero for every x, so y is at least 5 everywhere, with equality only at x equal to 2. That makes (2,5) the lowest point — which is what vertex means for an upward-opening V.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 125-125
Trap
\[ y = \lvert x + 4 \rvert - 2 \]
Read the plus four as a shift right by four
Why: The sign inside the bars is read as written, like the one outside.
\[ \text{vertex } (4, -2) \quad \text{(wrong)} \]
Substituting x equal to 4 gives the absolute value of 8, minus 2, which is 6 — not the vertex value of negative two.
\[ y = \lvert x + 4 \rvert - 2 = \lvert x - (-4) \rvert + (-2) \]
Find the value of x that makes the inside zero
Why: That value IS h, because the vertex is where the absolute value part contributes nothing.
\[ x + 4 = 0 \;\Longrightarrow\; x = -4 \;\Longrightarrow\; \text{vertex } (-4, -2) \]
Asking what makes the inside zero never requires remembering which way the sign flips.
Fill the middle
Rewrite to expose h, then read it off.
Fill in the blanks
y = \lvert x + 1 \rvert - 2 = \lvert x - (-1) \rvert + (-2) \;\Longrightarrow\; \text___ (-1, -2)
Why: Subtracting negative one is the same as adding one, so h is negative one. The vertex is therefore at (-1, -2). Substituting x equal to negative one confirms it: the inside is zero, so y is negative two, the lowest value the function reaches.
Discrimination
Do not graph anything. Say which direction.
Sort into buckets
Sort each equation by the direction of its horizontal shift.
Tweak it
Two dials, two independent directions.
Parameter explorer
Which of h and k moves the graph sideways, and which moves it up and down? Does either change the shape?
\[ y = \lvert x - {h} \rvert + {k} \]
Section
Section 3
Concept
In y equals a times the absolute value of x, the size of a decides the width and the sign of a decides which way the V opens. The vertex stays at the origin throughout.
\[ y = a\lvert x \rvert \]
When the size of a is greater than one the graph is stretched and looks narrower; when it is less than one the graph is shrunk and looks wider. A negative a reflects it in the horizontal axis.
Figure (svg): Three V graphs with different values of a: one wider than the parent, one narrower and inverted, and the parent itself
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 124-124 — Stretches, Shrinks, and Reflections
Picture it
Example 2: y equals one half the absolute value of x, and y equals negative three times it.
Figure (svg): Three V graphs with different values of a: one wider than the parent, one narrower and inverted, and the parent itself
Narrower and stretched are the same thing described two ways: the graph is pulled vertically, which makes it climb faster and therefore look thinner.
Worked example
Example 2a. The size of a is less than one.
\[ \text{Graph } y = \tfrac{1}{2}\lvert x \rvert \text{ and compare it with } y = \lvert x \rvert. \]
Read a and judge its size
Why: One half, whose size is less than one, so the graph is shrunk vertically and looks wider.
\[ a = \frac{1}{2},\text{ wider} \]
Note the vertex
Why: There is no h or k, so the vertex stays at the origin.
\[ \text{vertex } (0, 0) \]
Plot a point on each branch
Why: At x equal to 4 the value is 2; symmetry gives (-4, 2).
\[ (4, 2)\text{ and } (-4, 2) \]
Compare with the parent
Why: At x equal to 4 the parent gives 4 and this gives 2, so every height is halved.
Figure (svg): The solution to Worked example a shrink shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{2}\lvert x \rvert \quad \text{shrunk by a factor of } \tfrac{1}{2} \]
Verify: compare the branch slopes with the parent's
Why: The right branch runs from (0,0) to (4,2), a slope of one half rather than the parent's 1. Halving both slopes is exactly what multiplying the whole function by one half does, and a shallower slope is what wider means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 124-124
Sorting
Compare each with the parent V.
Sort into buckets
Sort each function by its width.
Width depends only on the SIZE of a. Direction depends only on its sign. Separating the two questions makes both easy.
Worked example
Example 2b. Both effects of a at once.
\[ \text{Graph } y = -3\lvert x \rvert \text{ and compare it with } y = \lvert x \rvert. \]
Read the size of a
Why: Three, which is greater than one, so the graph is stretched and looks narrower.
\[ | a | = 3,\text{ narrower} \]
Read the sign of a
Why: Negative, so the graph is reflected in the horizontal axis and opens downward.
Note the vertex
Why: Still the origin, since a does not move it.
\[ \text{vertex } (0, 0) \]
Plot a point on each branch
Why: At x equal to 1 the value is negative three; symmetry gives (-1, -3).
\[ (1, -3)\text{ and } (-1, -3) \]
Figure (svg): The solution to Worked example a stretch with a reflection shown as a ladder of expressions, one row per algebraic move
\[ y = -3\lvert x \rvert \quad \text{stretched by } 3 \text{ and reflected} \]
Verify: check the range
Why: Every value is three times a non-negative number, then negated, so y is at most zero and reaches zero only at the vertex. The graph lies on or below the horizontal axis, which is what a downward-opening V must do — and it is the mirror image of the parent's range.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 124-124
Trap
\[ y = 3\lvert x \rvert \]
Call it a shrink because the graph looks thinner
Why: Thin is read as small, and small is read as shrunk.
But the graph is thinner precisely because its values are three times LARGER at every input.
\[ y = 3\lvert x \rvert \]
Judge the stretch by what happens to the OUTPUT, not the width
Why: Multiplying by 3 triples every height, which is a vertical stretch. The narrowness is a consequence of the stretch, not a separate effect.
\[ \text{at } x = 2: \; \lvert x \rvert = 2 \text{ but } 3\lvert x \rvert = 6 \]
Size of a greater than one: stretched, and therefore narrower. Size less than one: shrunk, and therefore wider.
Prediction
Commit before reasoning.
Predict first
Which of these graphs opens downward?
Correct: y = -3 times the absolute value of x, only.
\[ y = 3\lvert x \rvert - 5 \;\text{ at }\; x = 10: \; y = 25 \quad \text{(rising: opens up)} \]
Why: The direction of opening is decided entirely by the sign of a, the coefficient in front of the bars. A minus sign somewhere else in the equation does not do it: y equals 3 times the absolute value of x, minus 5, has a positive a and simply sits five units lower, opening upward from the vertex (0,-5). Checking a single value settles it — at x equal to 10 that function gives 25, well above its vertex.
Comparison
Fill the blanks. Sign and size do different things.
Comparison matrix
| Value of a | Width | Direction |
|---|---|---|
| a = 3 | narrower | opens up |
| a = 1/2 | wider | opens up |
| a = -3 | narrower | opens down |
| a = -1/2 | wider | opens down |
| a = -1 | same as parent | opens down |
The two columns are decided by two independent features of one number. Reading the size and the sign as separate questions is what makes this table unnecessary once you have it.
Error analysis
A student describes the graph of y equals negative one half times the absolute value of x.
Annotate
On: \( y = -\tfrac{1}{2}\lvert x \rvert \;\Longrightarrow\; \text{narrower than the parent, opening upward} \)
Ask two separate questions of the coefficient: what is its size, and what is its sign. One answers width, the other answers direction.
Section
Section 4
Concept
The general form combines all three transformations. The vertex is at (h, k), the size of a sets the width, and the sign of a sets the direction. Reading those off the equation is the whole of the graphing.
\[ y = a\lvert x - h \rvert + k \]
Plot the vertex first, then use a as the slope of the right branch and its negative as the slope of the left one.
Figure (svg): A summary card showing the general form with each letter labelled by what it controls
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 124-125
Picture it
The summary card for the whole lesson.
Figure (svg): A summary card showing the general form with each letter labelled by what it controls
Only h has a sign that flips, and only because the form subtracts it. Asking what makes the inside zero sidesteps that entirely.
Worked example
Guided Practice 3. Read every letter before drawing.
\[ \text{Graph } f(x) = -3\lvert x + 1 \rvert - 2 \text{ and compare it with } y = \lvert x \rvert. \]
Find the vertex by setting the inside to zero
Why: x plus one is zero at x equal to negative one, and the constant outside is negative two.
\[ \text{vertex } (-1, -2) \]
Read the size of a
Why: Three, greater than one, so the graph is stretched and looks narrower.
Read the sign of a
Why: Negative, so the graph opens downward.
Plot the vertex and use a as the branch slope
Why: From (-1,-2), going one right and three down gives (0,-5); symmetry gives (-2,-5).
\[ (0, -5)\text{ and } (-2, -5) \]
Figure (svg): The solution to Worked example all three transformations shown as a ladder of expressions, one row per algebraic move
\[ f(x) = -3\lvert x + 1 \rvert - 2, \quad \text{vertex } (-1, -2) \]
Verify: substitute the vertex and one branch point
Why: At x equal to negative one the inside is zero, so f is negative two — the vertex. At x equal to zero the inside is 1, so f is negative three minus two, which is negative five, matching the plotted point. And since a is negative, negative two is the HIGHEST value the function reaches.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 125-125
Notation
Every symbol in this line does one job.
Annotate
On: \( y = a\lvert x - h \rvert + k \)
Three letters, three independent jobs. Nothing in the equation does two things at once, which is what makes the form worth reading before plotting.
Worked example
Combining a shrink with a translation.
\[ \text{Graph } y = \tfrac{1}{4}\lvert x - 4 \rvert + 1. \]
Find the vertex
Why: The inside is zero at x equal to 4, and the constant outside is 1.
\[ \text{vertex } (4, 1) \]
Read the size and sign of a
Why: One quarter: less than one, so wider; positive, so it opens upward.
Plot from the vertex using a as the branch slope
Why: Going four right and one up gives (8, 2); symmetry gives (0, 2).
\[ (8, 2)\text{ and } (0, 2) \]
Describe the whole transformation
Why: The parent V shrunk by one quarter, then slid right four and up one.
Figure (svg): The solution to Worked example a wide upward V, shifted shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{4}\lvert x - 4 \rvert + 1, \quad \text{vertex } (4, 1) \]
Verify: check the minimum value
Why: The absolute value part is at least zero, so y is at least 1 for every x, with equality only at x equal to 4. The vertex really is the lowest point, and at x equal to 0 the function gives one quarter of 4 plus 1, which is 2 — the plotted point.
Error analysis
A student graphs f of x equals negative three times the absolute value of x plus one, minus two.
Annotate
On: \( f(x) = -3\lvert x + 1 \rvert - 2 \;\Longrightarrow\; \text{vertex } (1, -2), \text{ opening up} \)
Two checks catch both errors at once: ask what makes the inside zero, and substitute one point either side of the vertex to see which way the graph goes.
Definition probe
Each described change comes from exactly one of the three.
Sort into buckets
Sort each effect by which letter causes it.
Fill the middle
Read all three letters.
Fill in the blanks
y = -2\lvert x - 3 \rvert + 7 \;\Longrightarrow\; \text3, 7 (___), \text___
Why: Setting x minus 3 to zero gives h equal to 3, and the constant outside gives k equal to 7, so the vertex is (3, 7). The coefficient is negative two: size two means narrower than the parent, and the negative sign means it opens downward, making 7 the maximum value rather than the minimum.
Ranking
Graphing y equals a times the absolute value of x minus h, plus k.
Put in order
Why: The vertex has to be plotted before the branches can be drawn from it, which fixes the first two steps. Reading a before stepping means you know both how steep and which way before you commit ink. The check at the end catches the two errors this section is built around: a mis-signed h and a missed reflection.
Section
Section 5
Concept
Reading a graph is the reverse of drawing one. The vertex hands you h and k directly, and substituting any other point on the graph gives an equation you can solve for a.
\[ y = a\lvert x - h \rvert + k \]
This is the same structure as Lesson 2.4's point-slope method: one feature places the graph, and one more point pins down its shape.
Figure (svg): A V-shaped path with vertex at 5 comma 8 passing through the origin, with the constant a solved for
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 125-125 — the hologram example
Picture it
Example 3: a beam path with vertex (5, 8) passing through the origin.
Figure (svg): A V-shaped path with vertex at 5 comma 8 passing through the origin, with the constant a solved for
The vertex was read straight off; only a needed solving for, and one substitution did it.
Worked example
Example 3. Vertex from the picture, a from a substitution.
\[ \text{A V-shaped path has vertex } (5, 8) \text{ and passes through } (0, 0). \text{ Write its equation.} \]
Write the general form with h and k filled in
Why: The vertex gives h equal to 5 and k equal to 8; only a is unknown.
\[ y = a | x - 5 | + 8 \]
Substitute the other known point
Why: Zero for y and zero for x.
\[ 0 = a | 0 - 5 | + 8 \]
Simplify the absolute value
Why: The absolute value of negative five is five.
\[ 0 = 5 a + 8 \]
Solve for a
Why: Subtracting 8 and dividing by 5 gives negative eight fifths.
\[ a = -\frac{8}{5} \]
Figure (svg): The solution to Worked example the reference beam path shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{8}{5}\lvert x - 5 \rvert + 8 \]
Verify: check both known points and the direction
Why: At x equal to 5 the equation gives 8, the vertex. At x equal to 0 it gives negative eight fifths times five plus eight, which is negative eight plus eight, or zero — the given point. And a is negative, so the graph opens downward with its peak at the vertex, which matches a beam that rises to a point and comes back down.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 125-125
Reverse engineer
A V has vertex (1, -3) and passes through (4, 3).
Fill in the blanks
3 = a\lvert 4 - 1 \rvert - 3 \;\Longrightarrow\; a = 2
Why: The absolute value of 4 minus 1 is 3, so the equation reads 3 equals 3a minus 3. Adding 3 gives 6 equals 3a, so a is 2. The graph therefore opens upward and is narrower than the parent, which fits: the given point sits above the vertex, so the graph must rise away from it.
Worked example
The same method when the vertex is a minimum.
\[ \text{A V has vertex } (-2, 1) \text{ and passes through } (2, 9). \text{ Write its equation.} \]
Fill in h and k from the vertex
Why: Negative two and one, so the inside of the bars is x plus two.
\[ y = a | x + 2 | + 1 \]
Substitute the other point
Why: Nine for y and two for x.
\[ 9 = a | 2 + 2 | + 1 \]
Simplify
Why: The absolute value of four is four, so the equation is 9 equals 4a plus 1.
\[ 9 = 4 a + 1 \]
Solve for a
Why: Subtracting one gives 8 equals 4a, so a is 2.
\[ a = 2 \]
Figure (svg): The solution to Worked example a graph opening upward shown as a ladder of expressions, one row per algebraic move
\[ y = 2\lvert x + 2 \rvert + 1 \]
Verify: check the point and the shape
Why: At x equal to 2 the equation gives 2 times 4 plus 1, which is 9 — the given point. And a is positive with size 2, so the graph opens upward and is narrower than the parent, meaning (-2, 1) is the lowest point. A graph through a point ABOVE its vertex must open upward, which the positive a confirms.
Error analysis
A student tries to find a for a graph with vertex (5, 8).
Annotate
On: \( y = a\lvert x - 5 \rvert + 8, \quad \text{substitute } (5, 8): \; 8 = a(0) + 8 \)
A second point only tells you something if it differs from the first in the way you are trying to measure. Here that means avoiding the vertex.
Real world
A ball bounces off the ground at the point (4, 0), rising to a peak of 5 metres at horizontal distance 4 and returning to the ground at distance 8. Treat the path as V-shaped, opening downward, with the peak as the vertex.
Discussion prompt
Write the equation of the path with vertex (4, 5) passing through (0, 0), and say what a means physically. Then say why an absolute value model is a poor description of a real bounce.
Hint: The path is symmetric about its peak, which is what makes an absolute value model tempting.
Answer:
\[ 0 = a\lvert 0 - 4 \rvert + 5 \;\Longrightarrow\; 0 = 4a + 5 \;\Longrightarrow\; a = -\tfrac{5}{4} \]
\[ y = -\tfrac{5}{4}\lvert x - 4 \rvert + 5 \]
The value of a is the slope of the descending branch: the ball falls 5 metres over 4 metres of horizontal travel. But a real trajectory is a smooth parabola, not a V — it has no sharp corner at the top, because a ball changes direction gradually. Chapter 4 gives the right model.
Elimination
A V has vertex (5, 8) and passes through the origin.
Eliminate the wrong options
Which equation describes it?
Survives elimination: A
Why: The vertex gives h equal to 5 and k equal to 8. Substituting the origin gives 0 equal to 5a plus 8, so a is negative eight fifths. Both known points check, and the negative a correctly opens the graph downward so that it can descend from the vertex to the origin.
Commit first
Answer, then rate your confidence honestly.
Predict first
How many points are needed to determine an absolute value function of the form y equals a times the absolute value of x minus h, plus k?
Correct: The vertex plus one other point.
\[ \text{vertex } (h,k) \;\to\; h, k \qquad \text{one more point} \;\to\; a \]
Why: There are three unknowns, a, h and k, so three pieces of information are needed — and the vertex supplies two of them at once, since it names both h and k. One further point then gives a single equation in a. Three arbitrary points would work in principle but would require solving a harder system, which is why the vertex is worth identifying first. And the other point must not be the vertex itself, since there a has no effect.
Comparison
Fill the blanks. Each letter does exactly one job.
Comparison matrix
| Letter | What it controls | How to read it |
|---|---|---|
| a, size | width | bigger than 1 narrower; less than 1 wider |
| a, sign | direction of opening | positive up, negative down |
| h | horizontal position | the x that makes the inside zero |
| k | vertical position | the constant outside, as written |
| h and k together | the vertex, at (h, k) | read both, plot one point |
Nothing in the table interacts with anything else. That independence is what makes the general form worth learning as a form rather than case by case.
Pattern
One routine graphs any absolute value function, and one reverses it.
Step one and the last half of step five are the two places this lesson goes wrong most: a mis-signed h, and a vertex substituted where it can tell you nothing.
OpenStax Algebra and Trigonometry 2e, §3.5 Transformation of Functions §3.5
Check
Find the vertex. Ask what makes the inside zero.
Check your understanding
What is the vertex of the graph of y = |x + 4| - 2?
Answer: A
Why: The inside is zero when x equals -4, so h is -4, and the constant outside is -2, so k is -2. Substituting x = -4 gives y = -2, confirming it.
Check
Describe the transformation. Size and sign separately.
Check your understanding
How does the graph of y = -3|x| compare with the graph of y = |x|?
Answer: A
Why: The size of a is 3, which exceeds 1, so the graph is stretched vertically and looks narrower. The sign is negative, so it is reflected in the horizontal axis and opens downward.
Check
Write the equation from a graph.
Check your understanding
A V-shaped graph has vertex (5, 8) and passes through (0, 0). What is its equation?
Answer: A
Why: The vertex gives h = 5 and k = 8. Substituting (0,0) gives 0 = 5a + 8, so a = -8/5. The negative a correctly opens the graph downward toward the origin.
Real world
A tolerance from Lesson 1.7: a bolt should be 12.00 mm across, and the cost of reworking one is 4 dollars for every 0.01 mm it deviates in either direction.
Discussion prompt
Write the rework cost as an absolute value function of the measured diameter, identify a, h and k, and say what each one means to the factory. What does the vertex represent?
Hint: A perfect bolt costs nothing to rework.
Answer:
\[ C = 400\lvert d - 12 \rvert \]
Here h is 12, the target diameter, and k is 0, because a bolt exactly on target needs no rework. The coefficient 400 is the cost in dollars per millimetre of deviation — four dollars per hundredth of a millimetre, scaled up.
The vertex (12, 0) is the ideal bolt: the single measurement that costs nothing. Everything else costs more, symmetrically, whether it is oversized or undersized — and that symmetry is exactly why an absolute value is the right model rather than a line.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the graph of y equals the absolute value of x, minus 3, the same as the graph of y equals the absolute value of the quantity x minus 3?
Correct: No — the first shifts down 3 and the second shifts right 3.
\[ \lvert x \rvert - 3: \; \text{vertex } (0,-3) \qquad \lvert x - 3 \rvert: \; \text{vertex } (3,0) \]
Why: Where the 3 sits relative to the bars decides everything. Outside the bars it is k, so it moves the graph vertically and the vertex goes to (0, -3). Inside the bars it is h, so it moves the graph horizontally and the vertex goes to (3, 0). The two graphs share nothing but their shape, and substituting x equal to 0 separates them at once: the first gives -3 and the second gives 3.
Explain it
They can graph lines but have never seen a V.
Discussion prompt
In four sentences or fewer, tell them how to graph y equals a times the absolute value of x minus h, plus k, without a table of values. Include the one trick that removes the sign confusion in h.
Hint: The trick is a question, not a rule.
Answer:
Start by finding the vertex: ask what value of x makes the expression inside the bars equal zero, and read the constant outside the bars as the height. Plot that point, then step one unit sideways and a units up or down to get the next point, and mirror it on the other side.
The trick is that question — what makes the inside zero — because it gives the horizontal position directly and never requires remembering that the sign flips.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the horizontal shift, ask what makes the inside zero rather than reading the sign. For stretch and shrink, compare the SIZE of a with one and remember that stretched means taller, hence narrower. For the flip, read the sign of a as a separate question from its size. For writing equations, take h and k from the vertex and substitute any other point for a. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
In the middle of a page draw one coordinate plane and graph the parent V, y equals the absolute value of x, marking its vertex and writing the equation of each branch beside it. Around it, draw four more small planes: one showing a graph shifted left and down, one shifted right and up, one stretched and reflected, and one shrunk. Label each with its equation and its vertex. Down one margin write the general form and draw an arrow from each of a, h and k to a one-line statement of what it controls, marking with a star the one whose sign flips. At the bottom of the page, invent a V-shaped graph by choosing a vertex and one other point, then work out its equation and check both points satisfy it.
If your starred letter is anything other than h, reread Section 2. Only the horizontal shift has a sign that reads backwards, and only because the form subtracts it.
Recap
Five things, and the fourth combines the other three into one reading.
| If you see | Then |
|---|---|
| Something added inside the bars | The graph shifts LEFT |
| Something added outside the bars | The graph shifts up |
| A coefficient bigger than 1 in size | Stretched, so narrower |
| A coefficient smaller than 1 in size | Shrunk, so wider |
| A negative coefficient | Opens downward |
Lesson 2.8 closes the chapter by shading regions instead of drawing lines: linear inequalities in two variables.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.7 Use Absolute Value Functions and Transformations §2.7, pp. 123-127 — everything on these slides traces back here
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