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2.9 — Lining Up Quadratics
You already know how to factor trinomials, graph parabolas, and read vertex form. Today's lesson connects these skills. Study each example, and reproduce the key graph points yourself before referring to the pictures.
1. In class: A Practice Understanding Task
Read the important points from factored form
A quadratic expressed as a product of two linear factors often looks like \(y=a(x-r)(x-s)\), where \(a\ne0\). The zero-product property says that a product is zero if at least one factor is zero. Because an x-intercept lies where \(y=0\), the zeros are \(x=r\) and \(x=s\). Write intercepts as points: \((r,0)\) and \((s,0)\).
Analyze \(y=(x-1)(x+3)\).
Zeros: \(x-1=0\Rightarrow x=1\); \(x+3=0\Rightarrow x=-3\). So the x-intercepts are \((1,0)\) and \((-3,0)\).
Axis: \(x=(1+(-3))/2=-1\).
Vertex: substitute \(x=-1\): \(y=(-2)(2)=-4\), hence \((-1,-4)\), a minimum.
y-intercept: when \(x=0\), \(y=(-1)(3)=-3\), hence \((0,-3)\).
Scale: \(a=1\); opens upward, neither vertically stretched nor compressed relative to \(y=x^2\).
Analyze \(y=2(x-2)(x-6)\).
The zeros are 2 and 6, so the axis is \(x=(2+6)/2=4\). At \(x=4\), \(y=2(2)(-2)=-8\), giving vertex \((4,-8)\). The y-intercept is \((0,2(-2)(-6))=(0,24)\). The parabola opens upward and its outputs are scaled by 2 relative to \((x-2)(x-6)\).
For \(y=-x(x+4)\), the x-intercepts are \((0,0)\) and \((-4,0)\); the axis is \(x=-2\); the vertex is \((-2,4)\) and is a maximum. Notice that the origin is both an x-intercept and the y-intercept. If both roots are the same, the axis passes through that single touching x-intercept.
Why the midpoint rule works
The graph is symmetric, so two points with the same y-value lie equally far to the left and right of the symmetry axis. Both roots have y-value zero. Thus the center must lie halfway between their x-coordinates. This is the same symmetry you saw earlier when graphing parabolas from tables.
A parabola from the product of two lines
Suppose two lines are \(L_1(x)=x+1\) and \(L_2(x)=x-3\). Multiplying their output values produces \(P(x)=L_1(x)L_2(x)=(x+1)(x-3)\), which is a quadratic function. Its graph is a parabola, not a straight line.
At \(x=-1\), the first line has output 0, so the product must be 0. At \(x=3\), the second line has output 0, so the product is again 0. The product has precisely the x-intercepts of the two factors (unless the factors share the same root).
At \(x=0\), the two lines have y-intercepts 1 and −3. The product's y-intercept is their product, \(1\cdot(-3)=-3\).
Let \(L_1(x)=2(x-1)\) and \(L_2(x)=-3(x+2)\). Then \(P(x)=-6(x-1)(x+2)\).
It crosses at \((1,0)\) and \((-2,0)\); the axis is \(x=-\tfrac12\). At \(x=-\tfrac12\), \(P(x)=-6(-\tfrac32)(\tfrac32)=\tfrac{27}{2}\), a maximum. At zero, the lines have y-intercepts −2 and −6, so the product's y-intercept is \((0,12)\). This matches substitution into the quadratic.
Working backward from a parabola: if its x-intercepts are \(r,s\), factors could be \(m(x-r)\) and \(\frac{a}{m}(x-s)\) for any nonzero \(m\). Their product is the same \(a(x-r)(x-s)\). Thus the individual factor lines are not unique, even when the parabola is fully specified.
Imagine a parabola crossing at \((-2,0)\) and \((4,0)\) with y-intercept \((0,-16)\). Begin with \(y=a(x+2)(x-4)\). At \(x=0\), \(-16=a(2)(-4)=-8a\), so \(a=2\). Therefore \(y=2(x+2)(x-4)\). One pair of factor lines is \(L_1(x)=2(x+2)\), \(L_2(x)=x-4\); another is \(L_1(x)=x+2\), \(L_2(x)=2(x-4)\).
The sketch indicates the intended zeros and general shape, but precise algebraic equations require readable coordinates and a scale point such as the y-intercept. Use the plotted x-intercepts to create \((x-r)(x-s)\); use a further point to determine \(a\). Multiple pairs of lines can have exactly the same product, but if all graph coordinates are specified exactly, the resulting quadratic function is unique.
2. Ready review — Multiply binomials using a two-way table
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.
To multiply two binomials, multiply every term in the first factor by every term in the second. A four-cell table keeps those products organized.
Expand \((3x-2)(2x+5)\).
| \(\times\) | \(2x\) | \(+5\) |
|---|---|---|
| \(3x\) | \(6x^2\) | \(15x\) |
| \(-2\) | \(-4x\) | \(-10\) |
Combine the four cells: \(6x^2+15x-4x-10=\boxed{6x^2+11x-10}\).
Two especially useful patterns come from products of matching or opposite binomials:
Expand \((4x-3)^2\). The first square is \(16x^2\); the two cross-products total \(-24x\); the last square is 9. Thus \(\boxed{16x^2-24x+9}\).
If we write the binomial as \((px+q)^2\), the coefficient of x in the expanded expression is \(2pq\), not \(p+q\). This is the pattern sought in Ready problem 10.
Check your work — Ready 1–10 answer key
- \(21 x^{2} - 43 x + 20\).
- \(9 x^{2} + 56 x + 12\).
- \(12 x^{2} + 35 x - 33\).
- \(49 x^{2} - 9\).
- \(9 x^{2} - 100\).
- \(121 x^{2} - 25\).
- \(16 x^{2} + 40 x + 25\).
- \(x^{2} + 18 x + 81\).
- \(100 x^{2} - 140 x + 49\).
- For standard coefficients \(ax^2+bx+c\), \(b^2=4ac\); the middle term is twice the product of the binomial terms, with its sign retained. Writing \((px+q)^2\) gives \(a=p^2,\ b=2pq,\ c=q^2\).
3. Set practice — Factored-form features
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.
Now apply the procedure to different signs and repeated factors. Be careful: the number written beside \(x\) inside a factor has the opposite sign from the zero.
For \(y=-\tfrac12(x-3)^2\): the only x-intercept is \((3,0)\); the axis is \(x=3\); the vertex is \((3,0)\), a maximum; the y-intercept is \((0,-\tfrac92)\); and \(a=-\tfrac12\) reflects downward and compresses the outputs by one half. A repeated zero touches the x-axis instead of crossing it.
For \(y=\tfrac13(x+1)(x-4)\), the zeros are −1 and 4, so the symmetry axis is \(x=\tfrac32\). Substituting gives \(y=\tfrac13(\tfrac52)(-\tfrac52)=-\tfrac{25}{12}\). The vertex is \((\tfrac32,-\tfrac{25}{12})\); the y-intercept is \((0,-\tfrac43)\). A fractional vertex coordinate is perfectly valid.
Check your work — Set 11–19 answer key
- (a) Vertex \(\left( -2, \ -64\right)\). (b) x-intercepts \(\left( -6, \ 0\right),\ \left( 2, \ 0\right)\). (c) y-intercept \(\left( 0, \ -48\right)\). (d) \(a=4\); opens up; vertical stretch by \(4\).
- (a) Vertex \(\left( 2, \ 48\right)\). (b) x-intercepts \(\left( -2, \ 0\right),\ \left( 6, \ 0\right)\). (c) y-intercept \(\left( 0, \ 36\right)\). (d) \(a=-3\); opens down; vertical stretch by \(3\).
- (a) Vertex \(\left( -6, \ -1\right)\). (b) x-intercepts \(\left( -7, \ 0\right),\ \left( -5, \ 0\right)\). (c) y-intercept \(\left( 0, \ 35\right)\). (d) \(a=1\); opens up; scale \(1\).
- (a) Vertex \(\left( 7, \ 0\right)\). (b) x-intercept \(\left( 7, \ 0\right)\). (c) y-intercept \(\left( 0, \ \frac{49}{2}\right)\). (d) \(a=\frac{1}{2}\); opens up; vertical compression by \(\frac{1}{2}\).
- (a) Vertex \((2,18)\). (b) x-intercepts \((-4,0),\ (8,0)\). (c) y-intercept \((0,16)\). (d) \(a=-\frac12\); opens down; vertical compression by \(\frac12\).
- (a) Vertex \(\left( 17, \ - \frac{192}{5}\right)\). (b) x-intercepts \(\left( 9, \ 0\right),\ \left( 25, \ 0\right)\). (c) y-intercept \(\left( 0, \ 135\right)\). (d) \(a=\frac{3}{5}\); opens up; vertical compression by \(\frac{3}{5}\).
- (a) Vertex \(\left( 0, \ - \frac{27}{4}\right)\). (b) x-intercepts \(\left( -3, \ 0\right),\ \left( 3, \ 0\right)\). (c) y-intercept \(\left( 0, \ - \frac{27}{4}\right)\). (d) \(a=\frac{3}{4}\); opens up; vertical compression by \(\frac{3}{4}\).
- (a) Vertex \(\left( 0, \ 25\right)\). (b) x-intercepts \(\left( -5, \ 0\right),\ \left( 5, \ 0\right)\). (c) y-intercept \(\left( 0, \ 25\right)\). (d) \(a=-1\); opens down; scale \(1\).
- (a) Vertex \(\left( -10, \ 0\right)\). (b) x-intercept \(\left( -10, \ 0\right)\). (c) y-intercept \(\left( 0, \ \frac{200}{3}\right)\). (d) \(a=\frac{2}{3}\); opens up; vertical compression by \(\frac{2}{3}\).
4. Go review — Recover features from vertex form
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 earlier sections you learned \(y=a(x-h)^2+k\). The vertex is \((h,k)\), and the axis is \(x=h\). The multiplier \(a\) still tells whether the parabola opens upward or downward and how steeply it changes. For the y-intercept, set \(x=0\). For the x-intercepts, set \(y=0\) and solve:
If \(-k/a\) is positive, there are two distinct real x-intercepts; if zero, one touching intercept; if negative, none. When positive, \(x=h\pm\sqrt{-k/a}\). The square root step is necessary unless the value is an obvious perfect square.
Find all features of \(y=(x+1)^2-9\).
Vertex: \((-1,-9)\). Axis: \(x=-1\). Scale: \(a=1\), opens upward. y-intercept: \(y=(0+1)^2-9=-8\), or \((0,-8)\). x-intercepts: \((x+1)^2=9\), so \(x+1=\pm3\), giving \((-4,0)\) and \((2,0)\).
Find all features of \(y=-2(x-1)^2+8\).
Vertex: \((1,8)\), a maximum. Axis: \(x=1\). Scale: \(a=-2\), downward reflection and vertical stretch by 2. y-intercept: \(-2(1)+8=6\), or \((0,6)\). x-intercepts: \(0=-2(x-1)^2+8\) means \((x-1)^2=4\), so \(x-1=\pm2\), hence \((-1,0)\) and \((3,0)\).
For \(y=3(x+2)^2+12\), the vertex is \((-2,12)\), the axis is \(x=-2\), and the scale is 3 upward. The y-intercept is \((0,24)\). To find x-intercepts, \(0=3(x+2)^2+12\) implies \((x+2)^2=-4\), impossible for real x. Therefore there are no real x-intercepts.
Proving two forms are equal
A factored expression and a vertex-form expression describe the same function when expansion makes them identical. For example,
\[4(x-2)(x+6)=4(x^2+4x-12)=4x^2+16x-48.\]
\[4(x+2)^2-64=4(x^2+4x+4)-64=4x^2+16x-48.\]
Both forms produce exactly the same quadratic. This two-sided expansion strategy is useful for Go problem 26; it does not depend on testing a few x-values.
The pairs in this exercise are meant to be the same functions, not merely graphs that happen to have matching vertices or intercepts. Expand each expression independently and compare the resulting terms, or perform valid algebra to transform one expression into the other.
Check your work — Go 20–26 answer key
- (a) Vertex \(\left( -2, \ -4\right)\). (b) x-intercepts \(\left( -4, \ 0\right),\ \left( 0, \ 0\right)\). (c) y-intercept \(\left( 0, \ 0\right)\). (d) \(a=1\); opens up; scale \(1\).
- (a) Vertex \(\left( -6, \ 3\right)\). (b) x-intercepts \(\left( -7, \ 0\right),\ \left( -5, \ 0\right)\). (c) y-intercept \(\left( 0, \ -105\right)\). (d) \(a=-3\); opens down; vertical stretch by \(3\).
- (a) Vertex \(\left( 1, \ -8\right)\). (b) x-intercepts \(\left( -1, \ 0\right),\ \left( 3, \ 0\right)\). (c) y-intercept \(\left( 0, \ -6\right)\). (d) \(a=2\); opens up; vertical stretch by \(2\).
- (a) Vertex \(\left( -2, \ -64\right)\). (b) x-intercepts \(\left( -6, \ 0\right),\ \left( 2, \ 0\right)\). (c) y-intercept \(\left( 0, \ -48\right)\). (d) \(a=4\); opens up; vertical stretch by \(4\).
- (a) Vertex \(\left( 2, \ 48\right)\). (b) x-intercepts \(\left( -2, \ 0\right),\ \left( 6, \ 0\right)\). (c) y-intercept \(\left( 0, \ 36\right)\). (d) \(a=-3\); opens down; vertical stretch by \(3\).
- (a) Vertex \(\left( -6, \ -1\right)\). (b) x-intercepts \(\left( -7, \ 0\right),\ \left( -5, \ 0\right)\). (c) y-intercept \(\left( 0, \ 35\right)\). (d) \(a=1\); opens up; scale \(1\).
- (a) Both sides expand to \(4x^2+16x-48\). (b) Both sides expand to \(-3x^2+12x+36\). (c) Both sides expand to \(x^2+12x+35\). Equal expansions prove each identity.
Additional Learning Resources
The links below were checked for relevant instructional content. For this day's lesson, concentrate on the features and graphing techniques rather than unrelated quadratic word problems.
| Resource | Supports | Scope / What to skip |
|---|---|---|
| Khan Academy — Graphing quadratics in factored form (video) | Roots, vertex, and graphs; main task and Set. | Close match. |
| Khan Academy — Graph quadratics in factored form (practice) | Independent practice locating graph features. | Close match. |
| Khan Academy — Graphing quadratics: vertex form (video) | Go review on reading vertices and intercepts. | Close match to Go, review rather than new main-task instruction. |
Teacher's review of the source packet
Main task 9–13: The instruction to sketch two factor lines from a parabola can be underdetermined when the picture is interpreted qualitatively or plotted coordinates are unclear. State which x-intercepts and y-intercept must be used; explain that rescaling the factors inversely gives infinitely many pairs of lines with the same product.
Ready 10: The letters a, b, c are used informally to describe binomial components and a middle coefficient; this can conflict with the conventional quadratic coefficients. Writing the binomial as \((px+q)^2\) makes the pattern unambiguous: the expanded x-coefficient is \(2pq\).
Coverage check — Section 2.9
| Packet component | Required skills | Where taught |
|---|---|---|
| Main 1–4 | Factored form, all graph features, scale | In class, worked examples 1–2 |
| Main 5–13 | Linear factor functions, their product, possible pairs of lines, reconstructing an equation | In class, worked examples 3–4, clarification |
| Ready 1–10 | Two-way multiplication tables; squares and conjugates; middle-term patterns | Section 2, worked examples 5–6 |
| Set 11–19 | Factored-form vertex, intercepts, vertical scaling including fractions and repeated roots | Section 3, worked examples 7–8 |
| Go 20–26 | Vertex-form features; solving for x-intercepts; proving factored and vertex forms equivalent | Section 4, original Go problems A–C and proof |
Source: Mathematics Vision Project, Secondary Math II, Module 2, §2.9, packet pp. 51–57. Math notation is rendered with MathJax from a CDN; an internet connection is required. All graphs are embedded SVG graphics.
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