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3.5 — Throwing an Interception
You already know that a parabola is symmetrical and that its x-intercepts occur wherever the function equals zero. When factoring is inconvenient, we can still find the intercepts exactly by working from the vertex. The distance to an intercept provides a useful bridge between graphing and algebra.
1. In class: A Develop Understanding Task
X-intercepts come in symmetric pairs
For a quadratic function, an x-intercept is a point \( (x,0) \) at which \(f(x)=0\). The axis of symmetry is the vertical line \(x=h\), passing through the vertex \((h,k)\). If the parabola crosses the x-axis at two points, the points have equal horizontal distance from this axis.
When the intercepts exist, write them as two separate points:
\[ (h-d,0) \qquad\text{and}\qquad (h+d,0), \]where \(d\) is the nonnegative horizontal distance from the axis \(x=h\) to either intercept. If \(d=0\), the points coincide.
Find the vertex, axis, and x-intercepts of \(f(x)=x^2-2x-3\).
Factor first: \(f(x)=(x-3)(x+1)\). Its roots are \(x=3\) and \(x=-1\); the axis lies halfway between them: \(h=(-1+3)/2=1\). The vertex height is \(f(1)=1-2-3=-4\), so the vertex is \((1,-4)\).
\[d=3-1=2,\quad h-d=-1,\quad h+d=3.\]Both intercepts are 4 units above the vertex, because their y-coordinate is zero and the vertex's is \(-4\).
Completing the square reveals the vertex
The expression \(x^2+bx\) becomes part of a perfect square when we add \((b/2)^2\). We use that fact without changing the function's value by adding and subtracting the same number.
Rewrite \(f(x)=x^2-6x+4\) in vertex form.
\[\begin{aligned}x^2-6x+4&=(x^2-6x+9)-9+4\\&=(x-3)^2-5.\end{aligned}\]Therefore the vertex is \((3,-5)\) and the axis is \(x=3\). For the intercepts, set the output to zero:
\[ (x-3)^2-5=0,\quad (x-3)^2=5,\quad x-3=\pm\sqrt5. \]\[\boxed{x=3-\sqrt5\quad\text{or}\quad x=3+\sqrt5}.\]The intercept distance is \(d=\sqrt5\), approximately 2.236. The intercepts are approximately \(0.764\) and \(5.236\).
The same method works with a vertical stretch
For \(f(x)=2x^2-8x+5\), factor 2 from the first two terms before completing the square:
\[\begin{aligned}2x^2-8x+5&=2(x^2-4x)+5\\&=2[(x-2)^2-4]+5\\&=2(x-2)^2-3.\end{aligned}\]The vertex is \((2,-3)\). Set the function equal to zero:
\[2(x-2)^2-3=0\quad\Longrightarrow\quad (x-2)^2=\frac32.\]\[\boxed{x=2\pm\sqrt{\frac32}=2\pm\frac{\sqrt6}{2}}.\]The distance is \(d=\sqrt{3/2}\), approximately 1.225. Therefore the x-intercepts are approximately \(0.775\) and \(3.225\).
Develop the quadratic formula, not just memorize it
Take the general quadratic \(f(x)=ax^2+bx+c\), with \(a\ne0\). We want to solve \(ax^2+bx+c=0\). First divide through by \(a\), which is allowed because \(a\) is nonzero.
When the numerator is nonnegative, take square roots on both sides:
\[\left|x+\frac b{2a}\right|=\frac{\sqrt{b^2-4ac}}{2|a|}.\]Choosing both signs and rearranging gives the equivalent familiar formula (the ± sign absorbs the sign of \(a\)):
\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}.\]Notice that \(h=-b/(2a)\), so this formula has the same center as the symmetry axis.
Steps: Write the equation equal to zero; identify signed coefficients \(a,b,c\); substitute into each expression; simplify the square root and fraction when possible; check with substitution or the graph.
Solve \(x^2-3x-1=0\). Here \(a=1\), \(b=-3\), and \(c=-1\).
\[\begin{aligned}x&=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-1)}}{2(1)}\\&=\frac{3\pm\sqrt{9+4}}2\\&=\boxed{\frac{3\pm\sqrt{13}}2}.\end{aligned}\]Approximations are \(x\approx-0.303\) and \(x\approx3.303\). Either can be checked by substitution; the exact answers use radicals.
When are there two real intercepts?
The quantity beneath the square root is \(b^2-4ac\), often called the discriminant. It tells whether the square root in the formula is real, and thus whether the parabola reaches the x-axis. This is a brief interpretation of today's formula; complex solutions belong to a later section.
| Value of \(b^2-4ac\) | What it means for the real x-intercepts |
|---|---|
| Positive | Two distinct real x-intercepts |
| Zero | One touching point (a repeated root) |
| Negative | No real x-intercepts |
Derive the same formula from symmetric intercepts
The packet's second derivation starts not by completing a square but by naming the axis \(x=h\) and its intercept distance \(d\). If real intercepts exist, the roots are \(h-d\) and \(h+d\). Thus
\[\begin{aligned}f(x)&=a[x-(h-d)][x-(h+d)]\\&=a[(x-h)+d][(x-h)-d]\\&=a[(x-h)^2-d^2]\\&=ax^2-2ahx+a(h^2-d^2).\end{aligned}\]Compare coefficients with \(ax^2+bx+c\):
Coefficient of \(x\):
\[b=-2ah.\]Constant term:
\[c=a(h^2-d^2).\]The first equation gives \(h=-b/(2a)\). Substitute that into the second:
\[\begin{aligned}d^2&=h^2-\frac ca\\&=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\&=\frac{b^2-4ac}{4a^2}.\end{aligned}\]Therefore, when the discriminant is nonnegative, \(d=\sqrt{b^2-4ac}/(2|a|)\). The two roots are \(h-d\) and \(h+d\), giving exactly the same pair as the quadratic formula. This derivation demonstrates why its solutions are balanced around \(-b/(2a)\).
Let \(f(x)=-x^2+4x+5\). Set it equal to zero, so \(a=-1\), \(b=4\), \(c=5\):
\[x=\frac{-4\pm\sqrt{4^2-4(-1)(5)}}{2(-1)}=\frac{-4\pm6}{-2}.\]The roots are \(x=-1\) and \(x=5\). The axis is \(x=2\), the vertex is \((2,9)\), and the graph opens downward. The formula still works because its derivation never requires \(a>0\).
These general derivations describe the location of real x-intercepts when they exist. Problem 7 begins with a factored form built from two intercepts, so its starting assumption requires \(b^2-4ac\ge0\). The quadratic formula itself still provides a useful algebraic expression when that quantity is negative, but such a parabola has no real x-intercepts.
2. READY — Convert units of area (Problems 1–6)
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.
Area measures two-dimensional space. If a length conversion has factor \(k\), the area conversion has factor \(k^2\). Because one yard is three feet, a square yard contains a \(3\times3\) array of square feet: \(1\,\mathrm{yd}^2=9\,\mathrm{ft}^2\). For reverse conversions, divide by the square conversion factor.
Convert \(11\,\mathrm{yd}^2\) to square feet:
\[11\,\mathrm{yd}^2\times\frac{9\,\mathrm{ft}^2}{1\,\mathrm{yd}^2}=\boxed{99\,\mathrm{ft}^2}.\]Multiplying by 3 instead of 9 would be a linear conversion, not an area conversion.
Convert \(288\,\mathrm{in}^2\) to square feet. Since \(1\,\mathrm{ft}=12\,\mathrm{in}\), we have \(1\,\mathrm{ft}^2=144\,\mathrm{in}^2\):
\[288\,\mathrm{in}^2\times\frac{1\,\mathrm{ft}^2}{144\,\mathrm{in}^2}=\boxed{2\,\mathrm{ft}^2}.\]Since \(1\,\mathrm{mile}=5280\,\mathrm{ft}\), one square mile is \(5280^2=27{,}878{,}400\,\mathrm{ft}^2\). A \(2\,\mathrm{mile}^2\) region contains \(55{,}756{,}800\,\mathrm{ft}^2\). Since \(1\,\mathrm m=100\,\mathrm{cm}\), \(3\,\mathrm m^2=3(100)^2=30{,}000\,\mathrm{cm}^2\).
Check your work — Ready 1–6 answer key
- \(63\text{ ft}^2\).
- \(720\text{ in}^2\).
- \(27878400\text{ ft}^2\).
- \(1000000\text{ cm}^2\).
- \(\frac{100}{3}\text{ yd}^2\).
- \(\frac23\text{ ft}^2\).
3. SET — Translate parabolas and track points (Problems 7–9)
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.
Recall the parent parabola \(f(x)=x^2\). Subtracting a number outside the square shifts every point downward; adding a number inside the parentheses shifts it left (subtracting shifts it right). Track a point's coordinates, not only the function's intercepts, as it moves.
Vertical shift down \(k\): \(y=f(x)-k\) sends \((x,y)\mapsto(x,y-k)\).
Horizontal shift left \(s\): \(y=f(x+s)\) sends \((x,y)\mapsto(x-s,y)\).
Horizontal shift right \(s\): \(y=f(x-s)\) sends \((x,y)\mapsto(x+s,y)\).
Compare \(f(x)=x^2\), \(g(x)=x^2-9\), and \(h(x)=(x+2)^2-9\).
The vertex moves \((0,0)\to(0,-9)\to(-2,-9)\). For \(g\), setting \(x^2-9=0\) gives x-intercepts \((-3,0)\) and \((3,0)\). These points correspond to \((-3,9)\) and \((3,9)\) on the parent \(f\), and to \((-5,0)\) and \((1,0)\) on \(h\). The points on \(h\) come from shifting those on \(g\) left 2 units.
Compare \(f(x)=x^2\), \(g(x)=x^2-4\), and \(h(x)=(x-1)^2-4\). The roots of \(g\) are \(-2,2\). After reversing the vertical shift, the corresponding points on \(f\) are \((-2,4)\) and \((2,4)\). Following the horizontal shift right 1 instead gives \((-1,0)\) and \((3,0)\) on \(h\).
That distinction matters: the original x-intercepts need not remain x-intercepts after a vertical shift, but a pure horizontal shift preserves their heights.
Check your work — Set 7–9 answer key
- a. Plot not shown. Vertices: \(f:(0,0),\ g:(0,-9),\ h:(-2,-9)\). b. All are congruent upward parabolas; \(h\) is \(g\) shifted 2 left. c. \(g\) is \(f\) shifted 9 down; \(h\) is \(f\) shifted 2 left and 9 down. d. \((-3,0),(3,0)\). e. On \(f\): \((-3,9),(3,9)\); on \(h\): \((-5,0),(1,0)\). Corresponding points move up 9 to \(f\) or left 2 to \(h\).
- a. Plot not shown. Vertices: \(f:(0,0),\ g:(0,-4),\ h:(1,-4)\). b. All are congruent upward parabolas; \(h\) is \(g\) shifted 1 right. c. \(g\) is \(f\) shifted 4 down; \(h\) is \(f\) shifted 1 right and 4 down. d. \((-2,0),(2,0)\). e. On \(f\): \((-2,4),(2,4)\); on \(h\): \((-1,0),(3,0)\). Corresponding points move up 4 to \(f\) or right 1 to \(h\).
- For \(y=(x-h)^2+k\), find points on \(y=x^2\) with output \(-k\) and shift them by \((h,k)\). Thus the x-intercepts are \((h\pm\sqrt{-k},0)\) when \(k\le0\); none when \(k>0\).
4. GO — Evaluate functions and solve for inputs (Problems 10–15)
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.
There are two different requests in these exercises. To evaluate \(f(2)\), insert \(x=2\) and calculate the output. To find where \(f(x)=16\), replace the entire output with 16 and solve for the input. The second request may produce two inputs, one, or none.
Given \(p(x)=(x-4)^2+1\), find \(p(0)\), \(p(4)\), the inputs for which \(p(x)=1\), and the inputs for which \(p(x)=10\).
\[p(0)=(-4)^2+1=17,\qquad p(4)=0^2+1=1.\]For \(p(x)=1\): \((x-4)^2=0\), so \(x=4\). For \(p(x)=10\): \((x-4)^2=9\), so \(x-4=\pm3\); thus \(x=1\) or \(7\).
Let \(q(x)=x^2+2x-7\). Find \(q(-3)\), the inputs giving zero, and the inputs giving output 8.
First, \(q(-3)=9-6-7=-4\). For zero output, solve \(x^2+2x-7=0\):
\[x=\frac{-2\pm\sqrt{4+28}}2=\boxed{-1\pm2\sqrt2}.\]For output 8, solve \(x^2+2x-15=0=(x+5)(x-3)\), giving \(x=-5\) or \(x=3\). Notice that the intercept equation and the output-8 equation are different.
Let \(r(x)=(x+2)^2+3\). Find the inputs giving output 2.
\[(x+2)^2+3=2\quad\Longrightarrow\quad (x+2)^2=-1.\]No real number has a negative square, so there is no real input with \(r(x)=2\). This matches the graph: the minimum output is 3. We do not need complex numbers for today's real-function questions.
Given \(s(x)=-(x+3)^2+12\), find \(s(0)\), \(s(-3)\), all real solutions of \(s(x)=0\), and all real solutions of \(s(x)=8\).
Substitute: \(s(0)=-9+12=3\), and \(s(-3)=12\). For zero: \((x+3)^2=12\), so \(x=-3\pm2\sqrt3\). For output 8: \((x+3)^2=4\), so \(x=-5\) or \(x=-1\).
Let \(t(x)=(x-6)(x+2)\). Find \(t(0)\), \(t(1)\), and where \(t(x)=0\).
Evaluations give \(t(0)=(-6)(2)=-12\) and \(t(1)=(-5)(3)=-15\). The zero-product property gives \(x=6\) or \(x=-2\). Those are the x-coordinates of the graph's intercepts.
For \(u(x)=2(x-1)^2-8\), find all \(x\) such that \(u(x)=10\).
\[2(x-1)^2-8=10\Longrightarrow (x-1)^2=9\Longrightarrow x-1=\pm3.\]Thus \(x=-2\) or \(x=4\). On the graph, these are the two intersections with the horizontal line \(y=10\).
“The points corresponding to the x-intercepts in \(g\)” means track the same geometric points under each transformation. On the parent graph they usually have positive y-coordinates and are not x-intercepts there. Under a horizontal shift, the corresponding points remain on the x-axis.
Check your work — Go 10–15 answer key
- a. −12; b. 0; c. \(x=-6,2\); d. \(x=-8,4\).
- a. 27; b. 2; c. No real solution; d. \(x=5\pm\sqrt{14}\).
- a. 9; b. 36; c. \(x=3\); d. \(x=-1,7\).
- a. 1; b. 6; c. \(x=2\pm\sqrt3\); d. \(x=2\).
- a. 25; b. 9; c. \(x=-5\); d. \(x=-8,-2\).
- a. 7; b. −1; c. \(x=-1\pm2\sqrt2\); d. \(x=-3,1\).
Additional learning resources
These resources supplement today's lesson; use only the indicated portion so you do not move ahead into later solving-method comparisons.
| Resource and direct link | Why use it | Scope / What to skip |
|---|---|---|
| Khan Academy — Quadratic formula proof review | Written derivation from completing the square, closely aligned to today's main task. | Close match. Focus on the algebraic derivation and interpretation of the two solutions. |
| Khan Academy — The quadratic formula (video) | A worked introduction to substituting coefficients into the formula. | Close match. Especially useful for main task Problem 6 and Go output equations. |
| Professor Leonard — Intermediate Algebra Lecture 11.2: Solving Quadratic Equations with the Quadratic Formula | Optional extended classroom-style lecture to reinforce the formula and its applications. | Some extra material. Use examples involving real quadratic roots and stop if the lecture discusses nonreal solutions or applications beyond this packet. No reliable timestamps verified. |
| Professor Leonard — Intermediate Algebra Lecture 11.1: Solving Quadratic Equations By Completing the Square | Optional longer explanation of the step used in the formula derivation. | Some extra material. Watch the portions on completing the square; no verified topic timestamps. |
Teacher's review — packet issues
No major mathematical errors were identified in the Section 3.5 main task. One wording issue is substantial enough to clarify for students: in Set Problems 7e and 8e, corresponding points on \(f\) are easy to mistake for x-intercepts; those points on the parent parabola generally have nonzero y-coordinate. Suggested wording: “Track the two points that are x-intercepts of \(g\) backward to \(f\), and forward to \(h\); give their coordinates on each graph.”
The formula derivation in main task Problem 7 assumes the quadratic has real intercepts; make that assumption explicit when introducing \(d\). The chapter's separate clarification states the limitation.
Curriculum coverage — Section 3.5 only
| Packet portion | Mathematics covered | Where taught |
|---|---|---|
| Main task 1–2 | Symmetry, vertex, x-intercepts, distance \(d\) | In class |
| Main task 3–4 | Vertex form, completing the square, exact intercepts with radicals | In class |
| Main task 5–6 | General quadratic formula and its use | In class |
| Main task 7–8 | Derivation from symmetric roots; downward-opening case | In class |
| Ready 1–6 | Squared unit conversion, including English and metric measures | Section 2 |
| Set 7–9 | Vertical/horizontal translations and matching corresponding coordinates | Section 3 |
| Go 10–15 | Evaluate functions; solve for inputs yielding specified outputs, including no real input | Section 4 and six original worked problems |
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