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2.2 — Transformers: More Than Meets the y’s
You should already know from Section 2.1 how to shift, reflect, and scale the parent parabola \(y=x^2\). Here we combine those changes and use the resulting equation to describe a graph precisely. Read the explanations and examples before attempting the corresponding packet questions.
1. In class: A Solidify Understanding Task
Combine familiar graph transformations
Begin with \(y=x^2\), whose vertex is \((0,0)\). To move a parabola right 2, change \(x\) to \(x-2\): \(y=(x-2)^2\). Multiplying the square by 2 makes it vertically steeper: \(y=2(x-2)^2\). Adding 1 moves the entire graph up one unit: \(y=2(x-2)^2+1\).
A square starts with side length \(x\). Decrease its side by 4 units, multiply its area by 3, then add 6 square units.
First, the new side length is \(x-4\), so its area is \((x-4)^2\). Tripling the area gives \(3(x-4)^2\). Adding 6 produces
\[A(x)=3(x-4)^2+6.\]As a square-area model, the side length must satisfy \(x\ge4\). If we instead graph the algebraic quadratic function, we ordinarily allow all real \(x\).
For \(a\ne0\), the vertex is \((h,k)\), the axis of symmetry is \(x=h\), and the graph opens up when \(a>0\) or down when \(a<0\). The vertical scale factor is \(|a|\). A value \(|a|>1\) looks narrower; \(0<|a|<1\) looks wider.
Why the vertex is \((h,k)\). Since a square never becomes negative, \((x-h)^2\) reaches its smallest value, zero, at \(x=h\). Therefore, \(y=k\) at that input. Positive \(a\) gives a minimum; negative \(a\) gives a maximum. Note the sign inside the parentheses: \((x+3)^2\) means \(h=-3\), not \(+3\).
Graph vertex form accurately by hand
A sketch should use actual coordinates, not just the direction of the opening. A useful pattern is to start at the vertex and evaluate inputs one and two steps on either side. For \(y=a(x-h)^2+k\), moving one unit left or right from \(h\) changes \(y\) by \(a\); moving two units changes it by \(4a\); moving three changes it by \(9a\).
First, read the vertex as \((-1,3)\). Its symmetry line is \(x=-1\), and the negative multiplier means the graph opens downward. Then compute symmetric values:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) |
|---|---|---|---|---|---|
| \(x+1\) | −2 | −1 | 0 | 1 | 2 |
| \(y=-2(x+1)^2+3\) | −5 | 1 | 3 | 1 | −5 |
Plot all five points, connect them with a smooth downward curve, and draw a dotted line at \(x=-1\). The pairs \((-2,1)\) and \((0,1)\) are equally far from that line; so are \((-3,-5)\) and \((1,-5)\).
Graph \(y=\tfrac12(x-3)^2-2\). The vertex is \((3,-2)\), the axis is \(x=3\), and the parabola opens up, wider than \(y=x^2\). At \(x=2,4\), \(y=-\tfrac32\); at \(x=1,5\), \(y=0\). Thus five useful points are \((1,0),(2,-1.5),(3,-2),(4,-1.5),(5,0)\).
Write an equation from a graph, table, or description
Start with the vertex to find \(h\) and \(k\). Then use one other point \((x,y)\) to find the multiplier \(a\) by substituting into \(y=a(x-h)^2+k\). Do not assume every parabola has \(a=1\).
A parabola opens upward, has vertex \((2,-3)\), and passes through \((3,1)\). Start with \(y=a(x-2)^2-3\). Substitute the other point:
\[1=a(3-2)^2-3 \quad\Rightarrow\quad 4=a.\]Therefore \(\boxed{y=4(x-2)^2-3}\). Check: at \(x=3\), it gives \(4-3=1\).
Suppose the pairs are \((-2,8),(-1,2),(0,0),(1,2),(2,8)\). The minimum is \((0,0)\), so write \(y=ax^2\). Since \((1,2)\) belongs to the graph, \(2=a(1)^2\); hence \(y=2x^2\). It also fits \((2,8)\), because \(2\cdot4=8\).
If the graph's vertex falls between integer grid lines: use symmetry. For example, two points at the same height with inputs \(x=2\) and \(x=3\) have their midpoint at \(x=2.5\). The symmetry line is \(x=2.5\), even if no table entry occurs there.
Does the order of transformations matter?
The location and scaling described by the final vertex-form equation are unambiguous, but changing the sequence of operations can change the result. In particular, a vertical scale and a vertical translation do not generally commute.
Starting with \(y=x^2\), first multiply the outputs by 2, then move up 3:
\[y=2x^2+3.\]If you first move up 3 and then multiply all outputs by 2, the result is
\[y=2(x^2+3)=2x^2+6.\]The two vertices are \((0,3)\) and \((0,6)\), so the transformations give different graphs.
Horizontal and vertical translations can be interchanged without changing the final graph; a horizontal translation and a vertical scaling can also be interchanged. But when a later vertical scaling acts on an earlier vertical shift, it also scales that shift. To interpret \(a(x-h)^2+k\) consistently, start with horizontal shift and vertical scaling/reflection, then add the vertical shift \(k\).
2. Ready review: Standard form and multiplying expressions
The packet's Ready problems introduce another way to write a quadratic expression. They ask you to recognize its coefficients and distribute products. This is not yet a method for finding a vertex from an arbitrary standard-form expression.
The coefficient \(a\) multiplies \(x^2\), \(b\) multiplies \(x\), and \(c\) is the constant. A missing term has coefficient zero: \(y=-4x^2+7\) has \(a=-4,b=0,c=7\).
For \(y=8-3x^2+5x\), reorder the terms as \(y=-3x^2+5x+8\). Thus \(a=-3\), \(b=5\), \(c=8\). The order of the written terms does not change their coefficients.
When multiplying parentheses, distribute every term in one factor across every term in the other. Combine like terms only after all products have been found.
So \(a=2,b=5,c=-12\). Check the middle term: \(+8x-3x=+5x\), not \(+11x\).
For the second line the negative sign applies to every term after the square is expanded. A square is not simply the sum of the squares: \((x+2)^2\ne x^2+4\).
The opposite middle terms cancel; the coefficient of \(x\) is zero.
Ready homework — Problems 1–12
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.
Review standard form and multiplying expressions above.
Check your work — Ready 1–12 answer key
- \(a=5,\ b=3,\ c=6\).
- \(a=1,\ b=-7,\ c=3\).
- \(a=-2,\ b=3,\ c=0\).
- \(a=6,\ b=0,\ c=-5\).
- \(a=-3,\ b=4,\ c=0\).
- \(a=8,\ b=-5,\ c=-2\).
- \(y=x^{2} - 4 x\); \(a=1,\ b=-4,\ c=0\).
- \(y=2 x^{2} - 3 x + 1\); \(a=2,\ b=-3,\ c=1\).
- \(y=9 x^{2} - 4\); \(a=9,\ b=0,\ c=-4\).
- \(y=x^{2} + 12 x + 36\); \(a=1,\ b=12,\ c=36\).
- \(y=x^{2} - 6 x + 9\); \(a=1,\ b=-6,\ c=9\).
- \(y=- x^{2} - 10 x - 25\); \(a=-1,\ b=-10,\ c=-25\).
3. Set review: How many x-intercepts?
An x-intercept is where a graph meets the horizontal axis, so \(y=0\). For a parabola, the vertex's height and opening direction tell you whether it crosses that axis twice, touches it once, or misses it entirely.
| Graph | Two intercepts | One intercept | Zero intercepts |
|---|---|---|---|
| Opens up | Vertex below the axis | Vertex on the axis | Vertex above the axis |
| Opens down | Vertex above the axis | Vertex on the axis | Vertex below the axis |
Find values of \(k\) so that \(y=x^2+k\) has two, one, or no x-intercepts. The vertex is \((0,k)\) and the graph opens upward. Choose \(k=-2\) for two intersections, \(k=0\) for one, and \(k=2\) for none. For \(y=-x^2+k\) the pattern reverses: \(k=2,0,-2\), respectively.
Set homework — Problems 13–25
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.
Review intercept counts and graphing vertex form above.
Check your work — Set 13–25 answer key
- (a) Vertex \((0,0)\). (b) For \(x=-3,-2,-1,0,1,2,3\), outputs are \(9,4,1,0,1,4,9\). Plot not shown.
- Example \(\omega=-1\). \(\omega\lt0\).
- Example \(\omega=0\). \(\omega=0\).
- Example \(\omega=1\). \(\omega\gt0\).
- Example \(\omega=1\). \(\omega\gt0\).
- Example \(\omega=0\). \(\omega=0\).
- Example \(\omega=-1\). \(\omega\lt0\).
- Vertex \(\left( 1, \ 0\right)\); axis \(x=1\); points \(\left( -2, \ 9\right),\ \left( -1, \ 4\right),\ \left( 0, \ 1\right),\ \left( 2, \ 1\right),\ \left( 3, \ 4\right),\ \left( 4, \ 9\right)\). Plot not shown.
- Vertex \(\left( 1, \ 1\right)\); axis \(x=1\); points \(\left( -2, \ 10\right),\ \left( -1, \ 5\right),\ \left( 0, \ 2\right),\ \left( 2, \ 2\right),\ \left( 3, \ 5\right),\ \left( 4, \ 10\right)\). Plot not shown.
- Vertex \(\left( 1, \ 1\right)\); axis \(x=1\); points \(\left( -2, \ 19\right),\ \left( -1, \ 9\right),\ \left( 0, \ 3\right),\ \left( 2, \ 3\right),\ \left( 3, \ 9\right),\ \left( 4, \ 19\right)\). Plot not shown.
- Vertex \(\left( -3, \ 0\right)\); axis \(x=-3\); points \(\left( -6, \ 9\right),\ \left( -5, \ 4\right),\ \left( -4, \ 1\right),\ \left( -2, \ 1\right),\ \left( -1, \ 4\right),\ \left( 0, \ 9\right)\). Plot not shown.
- Vertex \(\left( -3, \ -4\right)\); axis \(x=-3\); points \(\left( -6, \ -13\right),\ \left( -5, \ -8\right),\ \left( -4, \ -5\right),\ \left( -2, \ -5\right),\ \left( -1, \ -8\right),\ \left( 0, \ -13\right)\). Plot not shown.
- Vertex \(\left( -1, \ 4\right)\); axis \(x=-1\); points \(\left( -4, \ - \frac{1}{2}\right),\ \left( -3, \ 2\right),\ \left( -2, \ \frac{7}{2}\right),\ \left( 0, \ \frac{7}{2}\right),\ \left( 1, \ 2\right),\ \left( 2, \ - \frac{1}{2}\right)\). Plot not shown.
4. Go review: Read the vertex, symmetry, and intercept count from a table
In the packet's Go questions, assume the displayed values come from quadratic functions. When outputs decrease and then increase, the vertex is a minimum; when they increase and then decrease, it is a maximum. Equal outputs at inputs equally spaced from a symmetry line are especially useful.
| \(x\) | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| \(y\) | 10 | 4 | 2 | 4 | 10 |
Outputs on opposite sides of \(x=1\) match. The vertex is \((1,2)\), axis \(x=1\), and it is a minimum. Because that minimum is above the x-axis, the parabola has zero x-intercepts.
| \(x\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| \(y\) | −2 | 2 | 2 | −2 |
Since the values at \(x=1\) and \(x=2\) agree, the symmetry line lies halfway: \(x=1.5\). A quadratic fitting the table is \(y=-2(x-1.5)^2+2.5\). Its vertex \((1.5,2.5)\) is a maximum, and because it is above the x-axis while the parabola opens down, the graph has two x-intercepts.
How did we get \(2.5\)? In \(y=a(x-1.5)^2+k\), the first differences \(+4,0,-4\) have constant second difference \(-4\), so \(a=-2\) for unit x-steps. Substituting \((1,2)\) gives \(2=-2(0.5)^2+k\), hence \(k=2.5\).
Quick check with differences. A quadratic table with equally spaced x-values has constant second differences. For example, outputs \(7,2,-1,-2,-1\) have first differences \(-5,-3,-1,+1\), and second differences \(+2,+2,+2\). Constant second differences support a quadratic model in these exercises, but a finite table alone cannot prove the function's behavior at every real input.
5. New, fully worked Go-style practice
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.
Assume the table is from a quadratic function. Find its vertex, axis, minimum or maximum, and number of x-intercepts.
| \(x\) | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| \(y\) | 13 | 4 | 1 | 4 | 13 |
Solution. The smallest output is 1 at \(x=0\), and the values match symmetrically on both sides. Vertex: \((0,1)\). Axis: \(x=0\). It is a minimum. Since the curve opens upward and its lowest point is above the x-axis, there are no x-intercepts.
Assume \((-2,-15),(-1,-3),(0,5),(1,9),(2,9),(3,5)\) come from a quadratic. Find the exact vertex, symmetry line, whether the vertex is a maximum or minimum, and the number of x-intercepts.
Solution, step 1: The equal values \(9\) at \(x=1,2\) show that the symmetry line is halfway: \(x=1.5\).
Step 2: The first differences are \(12,8,4,0,-4\), so second differences are all \(-4\). For unit-spaced inputs, \(2a=-4\), giving \(a=-2\).
Step 3: Use \((1,9)\) in \(y=-2(x-1.5)^2+k\):
\[9=-2(-0.5)^2+k=-0.5+k\quad\Rightarrow\quad k=9.5.\]Therefore the vertex is \(\boxed{(1.5,9.5)}\), its axis is \(\boxed{x=1.5}\), and it is a maximum. Since the parabola opens downward from a point above the x-axis, there are \(\boxed{2}\) x-intercepts. These may lie between table inputs; their exact locations are not required.
A quadratic is known to have outputs \(6,0,-2,0,6\) for inputs \(x=0,1,2,3,4\). Identify the vertex and number of x-intercepts, and explain whether the plotted data points alone would justify the conclusion without the quadratic assumption.
Solution. The symmetry is centered at \(x=2\). Vertex \((2,-2)\), axis \(x=2\), minimum. Because the parabola opens upward and the minimum is below zero, the graph intersects the x-axis twice (indeed at \(x=1,3\)). Without the quadratic assumption, the five values alone do not establish the number of roots or the function's behavior between and beyond those sampled inputs.
Check your work — Go 26–29 answer key
- (a) \((0,-6)\). (b) \(x=0\). (c) 2 x-intercepts. (d) Minimum.
- (a) \((2,1)\). (b) \(x=2\). (c) 0 x-intercepts. (d) Minimum.
- (a) \((-5,7)\). (b) \(x=-5\). (c) 2 x-intercepts. (d) Maximum.
- (a) \((-7,-8)\). (b) \(x=-7\). (c) 0 x-intercepts. (d) Maximum.
Packet clarifications — specific 2.2 exercises
The original word problem asks for the area when a side of length \(x\) is decreased by 3, the area is doubled, and 4 more square units are added. The expression is \(2(x-3)^2+4\). For a real square after shortening, \(x\ge3\). When examining the resulting function's entire parabola instead, mathematicians often use all real x-values. These are different domains for different purposes.
“Does it matter in which order?” cannot be answered with an unconditional yes or no. Some transformations commute, while others do not. The sequence matters whenever a later scaling affects an earlier translation, as demonstrated in Worked example 6.
The intended solutions treat the given tables as values of quadratic functions, but the directions do not state this. Assume a quadratic model for these exercises. A finite table cannot by itself prove a function is quadratic on its entire domain, even when second differences are constant.
The multiplier \(a\) produces a vertical scale factor \(|a|\), with a reflection across the x-axis when \(a\) is negative. Vertical scaling a translated parabola is not the same as uniform geometric scaling in both horizontal and vertical directions.
Additional learning resources
These externally hosted lessons were selected for Section 2.2 specifically. They require internet access.
| Resource and direct link | Why use it | Scope / What to skip |
|---|---|---|
| Khan Academy — Graphing quadratics: vertex form (video) | Short demonstration of graphing \(y=-2(x-2)^2+5\), closely matching the main task and Set graphing. | Close match. Watch the full video. |
| Khan Academy — Graph quadratics in vertex form (practice) | Interactive practice placing parabolas from their vertex-form equations. | Close match. Use after Section 2 of this supplement. |
| Khan Academy — Vertex-form graphing warmup | Quick practice finding key features and vertices. | Close match. Use for early confidence; proceed to graphing practice afterward. |
Teacher's review — Material issues only
Go problems 26–29 — Unstated global assumption. The instructions ask students to infer a parabola's vertex and global number of intercepts from a finite table, without explicitly asserting that each table represents a quadratic function. Different nonquadratic functions could agree at all listed x-values yet differ between them or elsewhere. Suggested correction: “Assume each table shows values from a quadratic function defined for all real x.”
Main task problem 9 — Underspecified order question. The question can be answered both ways, depending on which transformations are compared. Suggested correction: “Identify pairs of transformations that can be performed in either order; give an example where changing their order changes the final graph.”
No additional serious mathematical defects are identified in this section.
Coverage map — MVP 2.2 only
| Packet component | Skills needed | Where taught above |
|---|---|---|
| Main task 1–4 | Translate descriptions, graphs and tables into vertex form | In class and In class; examples 1, 4, 5 |
| Main task 5–8 | Graph; identify vertex, symmetry, reflection and dilation | In class; examples 2–3 |
| Main task 9 | Order of transformations | In class; example 6 |
| Ready 1–6 | Identify \(a,b,c\) in standard form | Section 2; example 7 |
| Ready 7–12 | Expand products and squared binomials; identify coefficients | Section 2; examples 8–10 |
| Set 13 | Plot parent \(y=x^2\), its vertex and symmetric coordinates | In class |
| Set 14–19 | Choose shifts for 0, 1, or 2 x-intercepts | Section 3; example 11; two graphs |
| Set 20–25 | Graph vertex form, including negative/fractional multipliers | In class; examples 2–3 |
| Go 26–29 | Read quadratic-table vertex, AoS, intercept count, max/min | Section 4 and Section 5; examples 12–13 and new A–C |
Prepared as a standalone companion to Mathematics Vision Project, Secondary Math II, Module 2, Section 2.2, “Transformers: More Than Meets the y’s” (curriculum CC BY 4.0). Graphs generated specifically for this supplement. Mathematical notation requires the externally hosted MathJax script when opening the file in a browser.
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