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2.5 — Be There or Be Square
You have already learned that a perfect square such as \((x+4)^2\) expands to \(x^2+8x+16\), and that the form \(y=a(x-h)^2+k\) makes a parabola's vertex easy to see. Today you will connect these ideas in the reverse direction: starting with an expanded expression and rewriting it without changing its value.
1. In class: A Practice Understanding Task
From expanded form to a perfect square
Consider \(y=x^2-6x+9\). The middle coefficient is \(-6\), so half is \(-3\). Squaring that number gives \(9\), exactly the constant term already present. Therefore
\[x^2-6x+9=(x-3)^2.\]Both expressions give the same output for every input. The right-hand form shows that the lowest point is \((3,0)\).
What if the constant does not complete the square?
For \(x^2+4x+2\), half of \(4\) is \(2\), so the square needs \(4\), not \(2\). Add and subtract the same quantity to keep the expression unchanged:
Check by expanding: \((x+2)^2-2=x^2+4x+4-2=x^2+4x+2\). Its vertex is \((-2,-2)\), and it opens upward.
Why does completing the square work visually?
The pieces \(x^2+6x\) can be pictured as an \(x\)-by-\(x\) square plus two strips, each \(x\)-by-3. They leave a missing 3-by-3 corner. Adding its area, \(9\), completes an \((x+3)\)-by-\((x+3)\) square.
When the coefficient of x is odd
Rewrite \(f(x)=x^2+7x+10\) in vertex form.
Half of 7 is \(7/2\), and \((7/2)^2=49/4\).
\[\begin{aligned}f(x)&=x^2+7x+\frac{49}{4}-\frac{49}{4}+10\\&=\left(x+\frac72\right)^2-\frac{49}{4}+\frac{40}{4}\\&=\left(x+\frac72\right)^2-\frac94.\end{aligned}\]Vertex: \((-7/2,-9/4)\). Do not round the fractions; exact numbers are useful for identifying the vertex and checking work.
When the coefficient of x² is not 1
Convert \(y=2x^2-12x+17\).
\[\begin{aligned}y&=2(x^2-6x)+17\\&=2\big[(x^2-6x+9)-9\big]+17\\&=2(x-3)^2-18+17\\&=2(x-3)^2-1.\end{aligned}\]Vertex: \((3,-1)\). The graph opens up, and its vertical stretch factor is 2. The added 9 inside the parentheses is multiplied by 2, so 18 must be subtracted outside.
Graph accurately from the rewritten equation
Once the equation reads \(y=a(x-h)^2+k\), mark the vertex \((h,k)\), draw the dotted symmetry line \(x=h\), and choose x-values equal distances from h. Calculate outputs in pairs. The sign of a determines up or down, and its magnitude determines width relative to \(y=x^2\).
Rewrite and graph \(y=x^2-4x+1\).
\[y=x^2-4x+4-4+1=(x-2)^2-3.\]Vertex \((2,-3)\); axis \(x=2\); minimum \(-3\). At x-values 2, 1/3, 0/4, and −1/5, the y-values are −3, −2, 1, and 6 respectively.
| x | −1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| y | 6 | 1 | −2 | −3 | −2 | 1 | 6 |
What makes a parabola look wider?
Compare \(y=(x+1)^2+4\) with \(y=\tfrac12(x-1)^2+\tfrac92\). The second graph changes y by only half as much for the same horizontal distance from its vertex; therefore it appears wider. A shift alone never changes width.
2. Ready review — Recognize a quadratic equation
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.
The first six homework problems ask whether an equation is quadratic. Simplify both sides first. An equation is quadratic in x when, after collecting like terms, its highest surviving power of x is 2. If the x² terms cancel completely, the equation is not quadratic.
This is a linear equation, not quadratic, even though x² appeared at first.
Compare \(3x(x-4)=0\): distributing gives \(3x^2-12x=0\), which is quadratic.
The prompt says to identify whether each equation represents a quadratic function, but several are equalities such as \(x^2+13x-4=0\), rather than functions written as \(y=f(x)\). Interpret the task as asking whether the equation is quadratic in x after simplification; an associated function can be made by setting \(y\) equal to the resulting quadratic expression.
Check your work — Ready 1–6 answer key
- Quadratic: \(x^2+13x-4\) has degree 2.
- Not quadratic: the \(3x^2\) terms cancel, leaving \(x=-2\).
- Quadratic: expanding gives \(4x^2-5x=0\).
- Not quadratic: simplifies to \(8x-17=0\), degree 1.
- Not quadratic: \(2^x\) is exponential, with the variable in the exponent.
- Quadratic: rewrite as \(4x^2-32=0\).
3. Set review — Complete squares with fractions
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.
The Set exercises extend today’s main technique to halves, thirds, fifths, and fourths. Use the same rule, with fraction arithmetic: \(x^2+bx+(b/2)^2=(x+b/2)^2\).
Complete the trinomial \(x^2-\tfrac35x+\square\).
\[\frac{-3/5}{2}=-\frac3{10},\qquad\left(-\frac3{10}\right)^2=\frac9{100}.\] \[x^2-\frac35x+\frac9{100}=\left(x-\frac3{10}\right)^2.\]For the homework, keep exact fractions rather than decimal approximations.
Convert \(y=-x^2+2x-5\) to vertex form.
\[\begin{aligned}y&=-(x^2-2x)-5\\&=-[(x-1)^2-1]-5\\&=-(x-1)^2-4.\end{aligned}\]The vertex \((1,-4)\) is a maximum, because the parabola opens downward.
Compare the leading multipliers after rewriting problems 9 and 10: problem 9 has \(a=1\); problem 10 has \(a=\tfrac12\). Thus problem 10 is wider. The graph’s apparent width should be compared at a common scale on both coordinate grids.
Check your work — Set 7–17 answer key
- \(y=\left(x - 2\right)^{2} - 3\); Vertex \(\left( 2, \ -3\right)\); axis \(x=2\); points \(\left( -1, \ 6\right),\ \left( 0, \ 1\right),\ \left( 1, \ -2\right),\ \left( 3, \ -2\right),\ \left( 4, \ 1\right),\ \left( 5, \ 6\right)\). Plot not shown.
- \(y=\left(x + 1\right)^{2} + 4\); Vertex \(\left( -1, \ 4\right)\); axis \(x=-1\); points \(\left( -4, \ 13\right),\ \left( -3, \ 8\right),\ \left( -2, \ 5\right),\ \left( 0, \ 5\right),\ \left( 1, \ 8\right),\ \left( 2, \ 13\right)\). Plot not shown.
- \(y=\left(x + \frac{3}{2}\right)^{2} + 1\); Vertex \(\left( - \frac{3}{2}, \ 1\right)\); axis \(x=- \frac{3}{2}\); points \(\left( - \frac{9}{2}, \ 10\right),\ \left( - \frac{7}{2}, \ 5\right),\ \left( - \frac{5}{2}, \ 2\right),\ \left( - \frac{1}{2}, \ 2\right),\ \left( \frac{1}{2}, \ 5\right),\ \left( \frac{3}{2}, \ 10\right)\). Plot not shown.
- \(y=\frac{\left(x - 1\right)^{2}}{2} + \frac{9}{2}\); Vertex \(\left( 1, \ \frac{9}{2}\right)\); axis \(x=1\); points \(\left( -2, \ 9\right),\ \left( -1, \ \frac{13}{2}\right),\ \left( 0, \ 5\right),\ \left( 2, \ 5\right),\ \left( 3, \ \frac{13}{2}\right),\ \left( 4, \ 9\right)\). Plot not shown.
- #10 is wider: \(a=\frac{1}{2}\) halves vertical distances from the vertex compared with \(a=1\).
- \(\frac{121}{4}\); \((x-\frac{11}{2})^2\).
- \(\frac{49}{4}\); \((x+\frac{7}{2})^2\).
- \(\frac{225}{4}\); \((x+\frac{15}{2})^2\).
- \(\frac{1}{9}\); \((x+\frac{1}{3})^2\).
- \(\frac{1}{100}\); \((x-\frac{1}{10})^2\).
- \(\frac{9}{64}\); \((x-\frac{3}{8})^2\).
4. Go review — Linear and quadratic recursive rules
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.
In a table with consecutive integer x-values, the first differences are changes between adjacent outputs. The second differences are the differences between those first differences. Equal first differences indicate the linear pattern intended in these exercises; equal, nonzero second differences indicate the quadratic pattern intended here.
A recursive rule gives a starting value and tells how to calculate the next output from the preceding one. An explicit rule calculates a value directly from its input, without needing the previous output.
For x=1,2,3,4,5, suppose f(x)=4,9,14,19,24.
First differences: +5, +5, +5, +5. The rule is linear, with slope 5. Since f(1)=4, the explicit equation is \(f(x)=5x-1\).
\[\boxed{f(1)=4,\quad f(n)=f(n-1)+5\text{ for }n\ge2.}\]Both the initial value and update rule are required for a complete recursive description.
For x=1,2,3,4,5, suppose f(x)=3,8,15,24,35.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f(x) | 3 | 8 | 15 | 24 | 35 |
| First differences | — | 5 | 7 | 9 | 11 |
| Second differences | — | — | 2 | 2 | 2 |
The step from input n−1 to n is \(2n+1\): at n=2 it is 5; at n=3 it is 7. Thus
\[\boxed{f(1)=3,\quad f(n)=f(n-1)+(2n+1)\text{ for }n\ge2.}\]Check: \(f(3)=f(2)+7=8+7=15\). These exercises ask for the recursive rule only when the pattern is quadratic.
Let g(1)=12, g(2)=17, g(3)=20, g(4)=21, g(5)=20.
First differences: +5, +3, +1, −1. Second differences: −2, −2, −2, so the pattern is quadratic and opens downward if extended as a quadratic function.
The step into position n is \(9-2n\) (for n=2 it is 5). Therefore
\[\boxed{g(1)=12,\quad g(n)=g(n-1)+(9-2n)\text{ for }n\ge2.}\]How to construct a recursive rule, step by step
- List outputs in increasing input order; check equal input spacing (the packet uses increments of 1).
- Subtract consecutive outputs to obtain first differences.
- If they are constant, state the initial output, constant-step recursive rule, and explicit linear rule.
- If first differences change by a constant nonzero amount, express the first difference as a linear function of n, and state the starting value plus the recurrence.
- Check two later outputs using the rule.
Assume each table is generated by a linear or quadratic function, with domain inputs continuing at integer steps. For a quadratic example, a valid response is an initial value such as \(f(1)=7\) and a recurrence valid for \(n\ge2\). An equation giving only the changing differences is incomplete without the starting value.
Check your work — Go 18–21 answer key
- Linear: \(f(n)=3n-3\); \(f(n+1)=f(n)+3,\ f(1)=0\). Rules apply for integers \(n\geq1\).
- Quadratic: \(f(n+1)=f(n)+3n,\ f(1)=7\). Rules apply for integers \(n\geq1\).
- Linear: \(f(n)=2n+6\); \(f(n+1)=f(n)+2,\ f(1)=8\). Rules apply for integers \(n\geq1\).
- Quadratic: \(f(n+1)=f(n)+2n+10,\ f(1)=28\). Rules apply for integers \(n\geq1\).
Additional learning resources
| Resource and direct link | Why use it | Scope / what to skip |
|---|---|---|
| Khan Academy — Completing the square review | Step-by-step reading for transforming expanded expressions into perfect squares. | Some extra material: focus on rewriting expressions; equation-solving steps are optional, not required today. |
| Khan Academy — Completing the square (video) | Visual introduction to the missing constant and square pattern. | Some extra material: use the perfect-square construction; ignore later solving-equation techniques if unfamiliar. |
| Khan Academy — Completing the square (intermediate practice) | Checks whether you can rewrite quadratics, especially when the constant needs adjustment. | Close match: targeted practice for today’s Set work. |
| OpenStax — 9.2, Complete the Square of a Binomial Expression | Written examples using half the linear coefficient and perfect squares. | Significantly broader: use only the subsection named here; later equation solving is beyond this lesson. |
Teacher / parent review of the packet
Ready 1–6: The wording switches between “quadratic equations” and “quadratic functions” for equations in x set equal to constants, without giving a y-variable. A clearer prompt is “Simplify each equation and determine whether it is quadratic in x.”
Go 18–21: The instructions ask students to decide linear versus quadratic from a finite table. As written, this conclusion is not logically forced without specifying the intended model family. Add: “Assume each table is generated by a linear or quadratic rule on consecutive integer inputs.”
Other section 2.5 tasks have no additional serious defects requiring correction.
5.5 packet-to-lesson coverage
| Packet component | Problems | Where taught |
|---|---|---|
| Main task | 1–4: vertex form, matching squares, compensating constants | In class and Section 5; worked examples 1 and 4 |
| Main task | 5–6: odd/negative coefficients; 7–8: nonunit multipliers | Worked examples 2, 3, and 7 |
| Ready | 1–6: classify simplified equations | Section 2 |
| Set | 7–11: rewrite, vertices, graphs, widths | In class and Section 5 and Section 3 |
| Set | 12–17: complete perfect-square trinomials with fractional coefficients | Section 3, worked example 6 |
| Go | 18–21: linear explicit/recursive, quadratic recursive | Section 4, three original worked Go problems |
Based on MVP Secondary Math II, Module 2, section 2.5, Mathematics Vision Project (CC BY 4.0). Graphs are computer-generated; an internet connection is needed to load the MathJax math typesetter.
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