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2.1 — Transformers: Shifty y’s
Start at the beginning even if you recognize some of the graphs. Each transformation is explained individually before we combine them. The square-root review near the end is a separate skill needed for today’s Go homework.
1. In class: A Develop Understanding Task
The square function and its parabola
The area of a square with side length \(x\) is \(A(x)=x^2\). Since side lengths cannot be negative, this physical model has domain \(x\ge 0\). As a graphing function, however, we may use all real numbers for \(y=x^2\), including negative inputs. For example, \((-3)^2=9\), just as \(3^2=9\).
The most basic quadratic graph, \(y=x^2\), is called the parent function; its graph is a parabola. The lowest or highest point of a parabola is its vertex. A vertical line that divides it into mirror-image halves is the axis of symmetry.
| \(x\) | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| \(x^2\) | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
Suppose the points are \((-2,4),(-1,1),(0,0),(1,1),(2,4)\). The pair of values at \(x=-2\) and \(x=2\) match; so do those at \(-1\) and 1. The middle point \((0,0)\) is the minimum, and the vertical mirror line is \(x=0\). The graph opens upward.
Moving the whole parabola up or down
A vertical translation adds the same number to every output. Compare \(y=x^2\), \(y=x^2+3\), and \(y=x^2-2\).
| \(x\) | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| \(x^2\) | 4 | 1 | 0 | 1 | 4 |
| \(x^2+3\) | 7 | 4 | 3 | 4 | 7 |
| \(x^2-2\) | 2 | −1 | −2 | −1 | 2 |
Graph \(y=x^2-4\). Start with the parent points \((-2,4),(-1,1),(0,0),(1,1),(2,4)\). Subtract 4 from each output to obtain \((-2,0),(-1,-3),(0,-4),(1,-3),(2,0)\). The vertex is \((0,-4)\), the symmetry line is \(x=0\), and the graph crosses the \(x\)-axis twice.
Moving the parabola left or right
Now change the input before squaring. In \(y=(x-2)^2\), the squared part is zero when \(x=2\). So the lowest point occurs at \((2,0)\): the original parabola has moved right 2. Likewise, \(y=(x+3)^2\) is zero at \(x=-3\), moving the parabola left 3.
The sign seems backward because it is the value that makes the parentheses zero—not the visible sign alone—that determines the new middle of the graph.
| Input \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \((x-2)^2\) | 4 | 1 | 0 | 1 | 4 |
To graph \(y=(x+2)^2\), start with the parent points and subtract 2 from every \(x\)-coordinate: \((-4,4),(-3,1),(-2,0),(-1,1),(0,4)\). The new vertex is \((-2,0)\). This changes inputs, not outputs.
Stretching, compressing, and reflecting
Multiplying all outputs by \(a\) gives \(y=ax^2\). If \(a=3\), each \(y\)-value triples; at \(x=1\) we move from 1 to 3. The graph is narrower (a vertical stretch). If \(a=\tfrac14\), the heights are one quarter as large, making the graph wider (a vertical compression).
If \(a\) is negative, positive heights become negative: the parabola reflects across the \(x\)-axis. Its vertex becomes a maximum rather than a minimum.
| Multiplier | What happens? |
|---|---|
| \(a>1\) | Opens up; narrower than \(y=x^2\). |
| \(0<a<1\) | Opens up; wider. |
| \(a=-1\) | Same width; opens down. |
| \(a<-1\) | Opens down; narrower. |
| \(-1<a<0\) | Opens down; wider. |
Compare \(y=-\tfrac12x^2\) with \(y=x^2\). At \(x=-2,-1,0,1,2\), the new outputs are \(-2,-\tfrac12,0,-\tfrac12,-2\). Since the coefficient is negative, it opens down; because its absolute value is between 0 and 1, it is wider. The vertex remains \((0,0)\), now a maximum.
Connecting the changes — after learning each separately
We can now describe a parabola with a horizontal shift, a scale/reflection, and a vertical shift together:
Start with \(y=x^2\). Move right \(h\) (or left if \(h\) is negative), multiply the vertical distances from the vertex by \(|a|\) and reflect if \(a\) is negative, then move up \(k\) (or down if negative). The vertex is \((h,k)\); the symmetry line is \(x=h\).
For \(y=-(x-1)^2+2\), first shift \(y=x^2\) right one, reflect across the \(x\)-axis, then shift up two. The vertex is \((1,2)\), the symmetry line is \(x=1\), and the vertex is a maximum. At \(x=0,2\), the height is \(-1+2=1\); at \(x=-1,3\), the height is \(-4+2=-2\). Five points are \((-1,-2),(0,1),(1,2),(2,1),(3,-2)\).
2. Ready homework — identifying graph features (1–10)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For the shared ideas used in these exercises, revisit the In class explanation.
In problems 1–6, mark the vertex at the turning point, then draw the vertical axis of symmetry through it. If the vertex is \((p,q)\), the symmetry line is \(x=p\)—not \(y=q\). An upward-opening graph has a minimum; a downward-opening graph has a maximum.
Imagine a graph with equal heights at \(x=1\) and \(x=5\). The symmetry line is halfway between: \(x=(1+5)/2=3\). If the lowest plotted height at that line is \(-2\), the vertex is \((3,-2)\). For a grid graph, count the squares carefully; the midpoint of equal-height points can help when the vertex falls between grid lines.
A parabola can have 0, 1, or 2 distinct \(x\)-intercepts: zero if its entire graph stays above or below the \(x\)-axis; one if its vertex touches the axis; two if the graph passes through the axis on both sides of its vertex.
For \(y=x^2+2\), every square is nonnegative, so \(x^2+2\) is always positive: 0 intercepts. For \(y=-x^2\), the graph touches the axis only at \(x=0\): 1 intercept. For \(y=x^2-9\), it reaches 0 at \(x=-3\) and \(x=3\): 2 intercepts.
Check your work — Ready 1–10 answer key
- Vertex \((0,0)\); minimum; axis \(x=0\). Plot not shown.
- Vertex \((4,0)\); minimum; axis \(x=4\). Plot not shown.
- Vertex \((-1,-2)\); minimum; axis \(x=-1\). Plot not shown.
- Vertex \((-3,5)\); maximum; axis \(x=-3\). Plot not shown.
- Vertex \((1,3)\); maximum; axis \(x=1\). Plot not shown.
- Vertex \((\frac{3}{2},-3)\); minimum; axis \(x=\frac{3}{2}\). Plot not shown.
- For vertex \((h,k)\), the axis is \(x=h\).
- The axis lies halfway between equal-height points. In #6 the x-intercepts are \((-2,0),(5,0)\), giving \(x=\frac{-2+5}{2}=\frac{3}{2}\); the graph places the minimum at \(y=-3\). The same midpoint rule works in #1–5.
- Zero, one, or two distinct x-intercepts.
- Example \(y=x^2+1\) has minimum \((0,1)\) and stays above the x-axis. An upward parabola has none when its minimum is positive; a downward parabola has none when its maximum is negative. Plot/construction not shown.
3. Set homework — area models and shifted tables (11–18)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For the shared ideas used in these exercises, revisit the In class explanation.
Area models are equations drawn as regions
A square with side \(x\) has area \(x^2\). Four such squares have total area \(4x^2\); a square with side \(4x\) has area \((4x)^2=16x^2\). A square with side \(x+4\) has area \((x+4)^2\). A square of area \(x^2\) together with a separate 4-unit region represents \(x^2+4\) (not necessarily a larger square).
“Add 4 to the side length before finding the area” means \((x+4)^2\). “Find the area and then add 4 square units” means \(x^2+4\). The placement of parentheses records which operation happens first.
Recover an equation from a table
For a vertical shift, compare each table output with the parent \(x^2\) value at the same input. The difference is constant. For a horizontal shift, find the input where the output is smallest (for upward-opening parabolas), then look for symmetric values on either side.
For inputs \(-2,-1,0,1,2\), suppose outputs are \(8,5,4,5,8\). The parent outputs would be \(4,1,0,1,4\); each has increased by 4. Therefore \(g(x)=x^2+4\), with vertex \((0,4)\).
For inputs \(1,2,3,4,5\), suppose outputs are \(4,1,0,1,4\). The minimum 0 occurs at \(x=3\). The values mirror on either side of 3, so the function is \(g(x)=(x-3)^2\). It is the parent graph shifted right three units.
The printed equations distinguish \( (4x)^2 \) from \(4x^2\). The first represents one square whose side is quadrupled (area \(16x^2\)); the second represents four original squares (area \(4x^2\). Keep the parentheses visible when matching the diagrams.
Check your work — Set 11–18 answer key
- d.
- b.
- c.
- a.
- (a) \(g(x)=x^2-7\). (b) Plot not shown; use the supplied table. (c) Down 7. (d) the outside −7 causes this shift.
- (a) \(g(x)=x^2+2\). (b) Plot not shown; use the supplied table. (c) Up 2. (d) the outside +2 causes this shift.
- (a) \(g(x)=(x+1)^2\). (b) Plot not shown; use the supplied table. (c) Left 1. (d) the inside +1 causes this shift.
- (a) \(g(x)=(x-3)^2\). (b) Plot not shown; use the supplied table. (c) Right 3. (d) the inside −3 causes this shift.
4. Go homework — simplifying square roots (19–24)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
This review skill is separate from graph transformations. The radical \(\sqrt{N}\) means the nonnegative square root of \(N\). In particular, \(\sqrt{49}=7\), never \(-7\), because the radical symbol chooses the principal square root.
The absolute-value bars matter: \(\sqrt{(-5)^2}=\sqrt{25}=5=|-5|\), not \(-5\). You may remove the bars only if the quantity inside is known to be nonnegative.
Simplify \(\sqrt{(t+6)^2}\). The squared quantity is \(u=t+6\), so \(\sqrt{(t+6)^2}=|t+6|\). If additionally \(t\ge -6\), the result is \(t+6\); otherwise retain absolute value.
Simplify \(\sqrt{64m^4n^6}\). Rewrite as \(\sqrt{(8m^2n^3)^2}=|8m^2n^3|=8m^2|n^3|=8m^2|n|^3\). We can remove the bars around \(m^2\) because a square is nonnegative.
Simplify \(\sqrt{(7p-10)^2}\), then evaluate at \(p=1\). Symbolically the answer is \(|7p-10|\). At \(p=1\), it becomes \(|7-10|=3\). Removing the bars too early would incorrectly produce \(-3\).
Simplify \(\sqrt{25q^2(3r^2-s)^2}\). This is \(\sqrt{[5q(3r^2-s)]^2}=|5q(3r^2-s)|=5|q|\,|3r^2-s|\). The factors inside the bars can each be negative, so keep both unless assumptions about their signs are given.
Check your work — Go 19–24 answer key
- \(7|ab^3|\).
- \(|x+13|\).
- \(|x-16|\).
- \(|36x+25|\).
- \(|11x-7|\).
- \(3|m(2p^3-q)|\).
5. Packet-specific interpretation notes
For the quilt's actual square side, \(x\ge0\). When the packet changes to graphing \(y=x^2\), it explicitly switches to all real-number inputs. If the side is increased by 5, physically relevant inputs remain \(x\ge0\) for the original side; this should not be confused with graphing the transformed function over all real numbers.
For these tasks, use the stated comparison with \(f(x)=x^2\) and recognize a vertical or horizontal shift. A finite table by itself does not prove that a function must follow the same equation for every input.
Additional Learning Resources
The links below were selected specifically for the transformations taught today. Study the relevant parts only, then return to the assigned exercises.
| Resource and direct link | What it covers / why use it | Scope / what to skip |
|---|---|---|
| Paul’s Online Math Notes — Transformations, Section 4.6 | Written rules and worked graphs for vertical/horizontal shifts and reflections; useful for main task 4–8 and Set. | Some extra material. Use the vertical-shift, horizontal-shift, and reflection rules; nonquadratic square-root/cubic/absolute-value examples are optional, not assigned. |
| Khan Academy — Intro to parabola transformations (video) | Explains why changing \(x\) and \(y\) moves or scales a parabola. | Close match. Relevant to the main task; the combined equation at the end is a recap rather than the starting point. |
| Khan Academy — Shift parabolas (practice) | Interactive graph-to-equation practice for horizontal and vertical shifts. | Close match. Focus on shifted parent parabolas, not other quadratic-solving lessons. |
| Khan Academy — Scale & reflect parabolas (practice) | Practice identifying multipliers and reflections of \(y=x^2\). | Close match. Directly supports main task 7–8. |
| Paul’s — Transformations practice problems with solutions | Extra worked practice. In the numbered list, problem 4 uses a shifted quadratic. | Some extra material. Do problem 4; skip the nonquadratic examples if they are unfamiliar. |
Teacher / parent review — material issues only
Set, 15–18: The instructions ask students to identify an equation from a finite list of values; they imply the given data arise from transformations of \(x^2\). Explicitly state this modeling assumption, because finite samples cannot establish the formula uniquely over all real numbers.
Go, 19–24: If the expected solutions simplify \(\sqrt{u^2}\) to \(u\), the packet needs a sign restriction. The mathematically correct general answer is \(|u|\), with absolute values removable only when the relevant expression is known nonnegative.
Section 2.1 coverage map
| Packet part | Skills included | Where taught |
|---|---|---|
| Main task 1–3 | Area domain; side versus area changes; compare altered quadratic graphs | In class and Section 3 |
| Main task 4–6 | Vertical and horizontal shifts; tables and graphical reasoning | In class |
| Main task 7–8 | Negative multiplier, reflection, vertical stretch/compression | In class |
| Ready 1–10 | Vertex, symmetry, maximum/minimum, graph x-intercepts, midpoint strategy | In class and Section 2 |
| Set 11–14 | Matching area models to expressions | Section 3 |
| Set 15–18 | Inferring translated equations from tables and graphing them | In class and Section 3 |
| Go 19–24 | Principal square roots, absolute values, square root of a squared expression | Section 4 (four new worked examples) |
Curriculum: Mathematics Vision Project, Secondary Math II, Module 2, Section 2.1, packet pp. 1–6 (PDF pages 4–9). This supplement is independently written; figures are generated from the equations shown. MathJax typesetting loads from a CDN, so an internet connection is needed for beautifully formatted mathematical notation.
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