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Week 9 · Chapter

Chapter 9 — Quadratic Functions & Their Graphs

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 6 (solve quadratic equations by factoring, the square root property, completing the square, and the quadratic formula; use the discriminant to predict the number of real solutions; and analyze quadratic functions and their graphs, including the vertex, intercepts, and maximum or minimum value), Objective 3 (use function notation; determine the domain and range of a function; evaluate functions at numeric and algebraic inputs; and form sums, differences, products, quotients, and compositions of functions) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

An object launched from an 80-foot platform climbs for 2 seconds, stops for an instant at 144 feet, and comes back down. This week you learn to read that kind of rise-and-fall from a formula, before you draw a single point. The curve it makes is a parabola (a U-shaped curve that is the same on both sides of a middle line). Its top or bottom point is the vertex (the turning point, where the curve stops going one way and starts going the other). Once you can find that point, you can answer questions such as how high, how much, and how big.

Why you already care about this

Picture an object launched straight up from the edge of an 80-foot platform. It rises for 2 seconds, tops out at 144 feet, and hits the ground at 5 seconds. Those three numbers tell the whole story of the flight. The middle one, the peak, is the number a sports analyst, a fireworks crew, or a drone pilot cares about most.

The same shape shows up far from any launch. A gardener with 40 feet of fencing wants the biggest garden it can enclose. A small bakery wants the cookie price that brings in the most money, because a price that is too low earns little and a price that is too high scares buyers away. In each case the answer sits at the turning point of a parabola.

In Week 7 you solved quadratic equations. You found where x² − 6x + 5 equals 0. This week you look at the whole function, every output at once, and you find its lowest point, its mirror line, and where it crosses each axis.

By the end of this chapter you will be able to:

  • read the vertex, the axis of symmetry, and the opening direction from vertex form;
  • find the vertex from standard form with the vertex formula and check it by substitution;
  • say whether the vertex is a minimum or a maximum, and state the range;
  • find the y-intercept and the x-intercepts, and use the discriminant to predict how many x-intercepts there are;
  • convert a quadratic function between vertex form and standard form;
  • solve a maximum or minimum problem and say what the vertex means in its setting.

Carry this sentence with you: the vertex is the turning point — find it first.

Where we are

  • Already covered: Week 3 named functions and their range; Week 4 found intercepts; Week 6 multiplied and factored polynomials; Week 7 solved quadratic equations and read the discriminant.
  • This week: the quadratic function and its graph — the parabola, vertex form, standard form, the vertex formula, intercepts, converting between forms, and maximum and minimum problems.
  • How they connect: a quadratic function's x-intercepts are the Week 7 solutions, its y-intercept is the Week 4 idea, and its vertex is a Week 3 function value.
  • Where we're going next: Week 10, Polynomial & Rational Functions, reads graphs of higher-degree polynomials and fractions of polynomials from their formulas.

Each earlier week hands you one tool. Week 3 taught you that f(x) names one output for one input, and that the range is the set of all outputs a function produces. Week 4 taught you that a y-intercept is where a graph meets the y-axis, found by setting x = 0. Week 6 taught you to expand a square such as (x − 3)² by FOIL. Week 7 taught you to solve x² − 6x + 5 = 0 three ways and to read the discriminant b² − 4ac before you solve. This week puts all four tools on one graph.

1. Meet the parabola: a table that turns around

Think about a skateboarder dropping into a half-pipe. The board falls fast at first, slows near the bottom, rolls through the lowest point, and climbs the far side at the same speeds in reverse. The path down and the path up are mirror images. A parabola behaves the same way.

You can see that shape in a table before you see it on a graph. Take the week's parabola from Week 7, f(x) = x² − 6x + 5, and evaluate it at the whole numbers 0 through 6.

  • A quadratic function is a function that can be written f(x) = ax² + bx + c with a not equal to zero, so the highest power of the input is 2; its graph is a parabola.
  • A parabola is the U-shaped curve that is the graph of a quadratic function; it opens either upward or downward and is the same on both sides of a vertical mirror line.

Worked example — the week's parabola as a table. Evaluate f(x) = x² − 6x + 5 at x = 0, 1, 2, 3, 4, 5, 6.

  1. One value, every step. You substitute 2 for x: f(2) = 2² − 6(2) + 5 = 4 − 12 + 5 = −3.
  2. The rest the same way. f(0), f(1), f(2), f(3), f(4), f(5), f(6) are 5, 0, −3, −4, −3, 0, 5.
  3. Add them as a check. 5 + 0 + (−3) + (−4) + (−3) + 0 + 5 = 0. The seven values sum to 0.

In words: the outputs fall from 5 to −4, bottom out, and climb back to 5. A straight line could never do that, because a line changes by the same amount at every step.

Outputs of the week's parabola f(x) = x² − 6x + 5 for the inputs 0 through 6
Input x Output f(x)
0 5
1 0
2 −3
3 −4
4 −3
5 0
6 5

Misconception check. Students often think a graph that goes down and then up must be two separate lines joined at a corner. It is tempting because a table of seven whole numbers looks like a set of straight pieces. The cure: compute a value between two inputs. f(2.5) = 6.25 − 15 + 5 = −3.75, which sits below the straight segment from −3 to −4. A parabola is one smooth curve with no corner.

2. Find the turning point: the vertex and the axis of symmetry

Fold a sheet of paper down the middle and the two halves match. Every parabola has a fold line like that. Look at the table again: f(2) and f(4) are both −3, f(1) and f(5) are both 0, and f(0) and f(6) are both 5. Each pair sits the same distance from x = 3.

  • The vertex is the turning point of a parabola, written (h, k): the lowest point when the parabola opens up and the highest point when it opens down.
  • The axis of symmetry is the vertical line x = h through the vertex of a parabola; points the same horizontal distance from it on either side have equal outputs.

Worked example — reading the vertex from the table. Use the table of f(x) = x² − 6x + 5.

  1. Find the turn. The outputs stop falling and start rising at x = 3, where f(3) = −4.
  2. Name the vertex. The vertex is (3, −4).
  3. Name the mirror line. The axis of symmetry is x = 3.
  4. Test the mirror. x = 2 is one step left of 3, and x = 4 is one step right. Both give −3, so the pairing holds.

In words: the lowest point of this parabola is (3, −4), and the graph is a mirror image across the vertical line x = 3. If you know one side, you know the other.

Symmetry also saves work. If you know the point (0, 5) is on the graph, you know its mirror partner. 0 is 3 steps left of the axis, so the partner is 3 steps right, at x = 6, and f(6) = 5 agrees.

Misconception check. Students often write the axis of symmetry as a number, "3", or as a point, "(3, −4)". It is tempting because the vertex is a point and h is a number. The cure: the axis is a line, so it is an equation. Write it x = 3, the vertical line through every point whose x-value is 3.

3. Read vertex form: the vertex in plain sight

Some formulas hide the vertex, and one form puts it on display. Compare two descriptions of the same trip: "drive 3 miles east, then 4 miles south" tells you where you end up, while a long list of turns makes you work it out. Vertex form is the short description.

  • Vertex form is a quadratic function written f(x) = a(x − h)² + k, so the vertex (h, k) and the axis x = h can be read off directly, with h taken from inside the parentheses with its sign reversed.

Why does the vertex sit at (h, k)? The squared part, (x − h)², is never negative, because a square is never negative. It equals 0 at exactly one input, x = h. So at x = h the function's output is a(0) + k = k, and that is the turning point.

Worked example — shift right and up. Read the vertex of (x − 2)² + 3.

  1. Match the form. (x − 2)² + 3 has a = 1, h = 2, and k = 3.
  2. Name the vertex and axis. The vertex is (2, 3), and the axis of symmetry is x = 2.
  3. Check by substitution. At x = 2, (2 − 2)² + 3 = 0 + 3 = 3. It matches.

In words: this parabola's lowest point is (2, 3), two units right of the y-axis and three units up.

Worked example — shift left and down. Read the vertex of (x + 3)² − 5.

  1. Rewrite the plus as a minus. x + 3 is the same as x − (−3). So (x + 3)² − 5 = (x − (−3))² − 5.
  2. Match the form. Now h = −3 and k = −5.
  3. Name the vertex. The vertex is (−3, −5), and the axis is x = −3.
  4. Check by substitution. (−3 + 3)² − 5 = 0 − 5 = −5. It matches.

In words: the lowest point is (−3, −5), three units left and five units down.

Misconception check. Students often read (x + 3)² − 5 as vertex (3, −5). It is tempting because the 3 in the formula is positive. The cure: substitute your answer. (3 + 3)² − 5 = 36 − 5 = 31, which is far from −5, so 3 cannot be the vertex's x-value. The number inside the parentheses comes out with its sign reversed: x + 3 means h = −3.

4. Decide up or down: minimum, maximum, and range

A satellite dish and an umbrella are both curved, but one cups upward and one cups downward. The number a, in front of the square, decides which way a parabola cups. You can read it before you draw anything.

When a is positive, a times a square is never negative, so the outputs can only go up from k. The parabola opens up, and k is the smallest output. When a is negative, a times a square is never positive, so the outputs can only go down from k. The parabola opens down, and k is the largest output.

  • The minimum value is the smallest output a function reaches; for a parabola that opens up (a is positive) it is k, the y-value of the vertex.
  • The maximum value is the largest output a function reaches; for a parabola that opens down (a is negative) it is k, the y-value of the vertex.

Worked example — the week's parabola. Decide the direction, the extreme value, and the range of f(x) = x² − 6x + 5, whose vertex is (3, −4).

  1. Read a. a = 1 is positive, so the parabola opens up.
  2. Name the extreme value. −4 is the minimum value of f.
  3. State the range. Every output is −4 or more, so the range is [−4, ∞). The bracket means −4 is included, and infinity always gets a parenthesis.

In words: the week's parabola never goes below −4, and it reaches every output from −4 upward.

Worked example — a parabola that opens down. Take −x² + 4x + 1, which you will meet again in Section 5 with its vertex (2, 5).

  1. Read a. a = −1 is negative, so the parabola opens down.
  2. Name the extreme value. 5 is the maximum value, not a minimum.
  3. State the range. Every output is 5 or less, so the range is (−∞, 5].

In words: this parabola tops out at 5 and runs downward forever on both sides.

Misconception check. Students often call every vertex a minimum, because the first parabolas they meet open up. It is tempting to think "vertex" means "bottom". The cure: look at the sign of a first. If a < 0, the vertex is the top of the curve, and its y-value is a maximum.

5. Find the vertex from standard form

Most formulas you meet are not in vertex form. A launch, a price, or a fence usually arrives as ax² + bx + c, with the vertex hidden. There is a formula that finds it in two steps.

  • Standard form of a quadratic function is a quadratic function written f(x) = ax² + bx + c with a not equal to zero. In this form the y-intercept (0, c) can be read off, and the vertex can be found from a and b.
  • The vertex formula is the rule h = −b ⁄ (2a) that gives the x-value of a parabola's vertex from standard form; the y-value is then k = f(h).

The vertex formula is not a new fact to memorize blindly. It is the midpoint of the two quadratic-formula solutions from Week 7: the ± part pushes the two solutions equal distances left and right of −b ⁄ (2a), so the axis sits exactly between them.

Worked example — the week's parabola from standard form. Find the vertex of f(x) = x² − 6x + 5.

  1. Read a, b, and c with their signs. a = 1, b = −6, and c = 5. The minus sign belongs to b, so b is −6, not 6.
  2. Find h. h = −b ⁄ (2a) = −(−6) ⁄ (2 · 1) = 6 ⁄ 2 = 3.
  3. Find k. k = f(3) = 3² − 6(3) + 5 = 9 − 18 + 5 = −4.
  4. Name the vertex and axis. The vertex is (3, −4), and the axis of symmetry is x = 3.

In words: the formula lands on the same turning point the table showed in Section 2. Two methods, one vertex.

Worked example — a parabola that opens down. Find the vertex of −x² + 4x + 1.

  1. Read a, b, and c. a = −1, b = 4, and c = 1.
  2. Find h. h = −4 ⁄ (2 · (−1)) = −4 ⁄ (−2) = 2.
  3. Find k. k = −(2)² + 4(2) + 1 = −4 + 8 + 1 = 5.
  4. Decide. The vertex is (2, 5), and because a = −1 is negative, 5 is the maximum value.

In words: this parabola peaks at the point (2, 5). Notice the first term: −(2)² is −4, because the square happens before the minus sign is applied.

Misconception check. Students often drop a sign in −b ⁄ (2a) when b is already negative, and get h = −6 ⁄ 2 = −3 for the week's parabola. It is tempting because the formula starts with a minus sign and b carries one too. The cure: write −(−6) in full, with the parentheses, and then simplify it to 6. Then check: f(−3) = 9 + 18 + 5 = 32, nowhere near the table's lowest value.

In practice (your tools). Open the free Desmos graphing calculator and type y = x^2 - 6x + 5. Desmos marks the vertex with a gray dot; click it and Desmos labels it (3, −4). Then type y = -x^2 + 4x + 1 on the next line, click its top point, and read (2, 5). Your formula and the picture should always agree.

6. Find the intercepts and let the discriminant predict them

A road map marks where a highway crosses each border. The intercepts of a parabola are the same kind of landmark: where the curve crosses the y-axis and where it crosses the x-axis. You already know how to find each one from earlier weeks.

The y-intercept comes from setting x = 0. In standard form every term with an x vanishes, so the output is c, and the y-intercept is (0, c). The x-intercepts come from setting the output to 0, which turns the function into the Week 7 equation ax² + bx + c = 0.

Worked example — the intercepts of the week's parabola. Find every intercept of f(x) = x² − 6x + 5.

  1. The y-intercept. You set x = 0: f(0) = 0² − 6(0) + 5 = 5, so the y-intercept is (0, 5), the point (0, c).
  2. Predict the x-intercepts. Before you factor, you compute the discriminant: b² − 4ac = (−6)² − 4(1)(5) = 36 − 20 = 16. Since 16 > 0, the graph crosses the x-axis twice.
  3. Find them. x² − 6x + 5 = 0 factors as (x − 1)(x − 5) = 0, so x = 1 or x = 5. The x-intercepts are (1, 0) and (5, 0).
  4. Check. 1 − 6 + 5 = 0 and 25 − 30 + 5 = 0.
  5. Tie them to the axis. The axis sits halfway between the x-intercepts: (1 + 5) ⁄ 2 = 6 ⁄ 2 = 3.

In words: the week's parabola crosses the y-axis at 5 and the x-axis at 1 and 5, and its mirror line runs exactly between the two crossings.

The discriminant tells you the count before you solve. A positive discriminant means two x-intercepts. A discriminant of 0 means one, where the vertex sits on the x-axis. A negative discriminant means none.

Worked example — no x-intercepts. Take (x − 2)² + 3 from Section 3.

  1. Expand. (x − 2)² + 3 = x² − 4x + 4 + 3 = x² − 4x + 7.
  2. Compute the discriminant. (−4)² − 4(1)(7) = 16 − 28 = −12.
  3. Read it. −12 is negative, so there are no x-intercepts.
  4. Check with the picture. The parabola opens up and its lowest point, at height 3, is already above the x-axis. It can never come down to 0.

In words: the formula and the picture agree. A parabola that opens up from a vertex above the x-axis never touches it.

Worked example — exactly one x-intercept. The Week 7 equation 2x² − 4x + 2 = 0 had discriminant (−4)² − 4(2)(2) = 16 − 16 = 0. As a function, f(x) = 2x² − 4x + 2 = 2(x − 1)², with vertex (1, 0). The vertex sits on the x-axis, so the graph touches the x-axis at one point. It does not cross.

Misconception check. Students often read the y-intercept as the vertex's y-value, or as the first coefficient a. It is tempting because both are "the number on the y-axis side". The cure: set x = 0 and compute. For the week's parabola, f(0) = 5, while the vertex's y-value is −4 and a is 1. The y-intercept of ax² + bx + c is always c.

7. Convert between the forms

A recipe can be written as a list of steps or as a list of ingredients, and each version answers a different question fast. Vertex form tells you the vertex at a glance. Standard form tells you the y-intercept and the a, b, and c the discriminant needs. You can always turn one into the other.

Worked example — vertex form to standard form. Expand 2(x − 1)² − 3.

  1. Square the binomial first. (x − 1)² = x² − 2x + 1, by the Week 6 pattern (a − b)² = a² − 2ab + b².
  2. Distribute the 2. 2(x² − 2x + 1) = 2x² − 4x + 2.
  3. Combine the constants. 2x² − 4x + 2 − 3 = 2x² − 4x − 1.
  4. Check at x = 0. 2(0 − 1)² − 3 = 2 − 3 = −1, and 2(0)² − 4(0) − 1 = −1. Both forms agree.
  5. Check the vertex. From standard form, h = −(−4) ⁄ (2 · 2) = 4 ⁄ 4 = 1, which matches the vertex (1, −3) from vertex form.

In words: 2(x − 1)² − 3 and 2x² − 4x − 1 are the same function, written two ways.

Worked example — standard form to vertex form by completing the square. Rewrite x² − 6x + 5 in vertex form.

  1. Half of b, squared. Half of −6 is −3, and (−3)² = 9.
  2. Add 9 and subtract 9. Adding and subtracting the same number changes nothing: x² − 6x + 5 = (x² − 6x + 9) − 9 + 5.
  3. Fold the square. x² − 6x + 9 = (x − 3)², and −9 + 5 = −4. So x² − 6x + 5 = (x − 3)² − 4.
  4. Check at x = 0. (0 − 3)² − 4 = 9 − 4 = 5, which matches f(0) = 5.

In words: the week's parabola in vertex form is (x − 3)² − 4, and the vertex (3, −4) reads off directly. The same move works on −x² + 4x + 1, which becomes −(x − 2)² + 5.

Misconception check. Students often expand 2(x − 1)² − 3 as 2x² − 2 − 3 or as (2x − 2)² − 3. Both are tempting shortcuts, and both are wrong. The cure: square first, distribute second. (x − 1)² has three terms, not two, and the 2 multiplies all three.

In practice (your tools). In Desmos, type y = 2(x - 1)^2 - 3 and, on the next line, y = 2x^2 - 4x - 1. The two graphs land exactly on top of each other, which is the picture of "same function, two forms". Click the vertex of either one and read (1, −3).

8. Solve maximum and minimum problems

A question that asks "how high", "how much", or "how big" is often a vertex question in disguise. Here is the move: write the quantity as a quadratic function, find the vertex, and say what each coordinate means in the setting.

Worked example — the launched object. The height in feet t seconds after launch is h(t) = −16t² + 64t + 80. The 16 is half of gravity's pull of 32 feet per second per second. In this setting the letter h names the height function; it is a different use of the letter from the vertex's h.

  1. Read the table. The heights at t = 0, 1, 2, 3, 4, 5 are 80, 128, 144, 128, 80, 0. The matching heights at t = 1 and t = 3 (128 feet) already point to the axis t = 2.
  2. Decide up or down. You read a = −16, which is negative, so the parabola opens down and the vertex is the highest point.
  3. Find when it peaks. t = −64 ⁄ (2 · (−16)) = −64 ⁄ (−32) = 2 seconds.
  4. Find how high. h(2) = −16(2)² + 64(2) + 80 = −16(4) + 128 + 80 = −64 + 128 + 80 = 144 feet.
  5. Read the rest. In vertex form, h(t) = −16(t − 2)² + 144. The y-intercept (0, 80) is the 80-foot platform.

In words: the object reaches its greatest height, 144 feet, two seconds after launch. A caveat on the setting: this model ignores air resistance, so it describes an idealized launch, not every real throw.

Worked example — the largest garden. You have 40 feet of fencing for a rectangular garden. How wide should it be to hold the most area?

  1. Name the unknown. Let w be the width in feet.
  2. Find the length. Two widths and two lengths use 40 feet, so each length is (40 − 2w) ⁄ 2 = 20 − w.
  3. Write the area. A(w) = w(20 − w) = −w² + 20w.
  4. Decide up or down. a = −1 is negative, so the area has a maximum.
  5. Find the best width. w = −20 ⁄ (2 · (−1)) = −20 ⁄ (−2) = 10 feet.
  6. Find the area. A(10) = 10(20 − 10) = 10 · 10 = 100 square feet, a 10-foot by 10-foot square.
  7. Check. You check the fencing: 2(10) + 2(10) = 40 feet. By symmetry, A(8) = 8 · 12 = 96 and A(12) = 12 · 8 = 96, both less than 100.

In words: with 40 feet of fencing, the biggest garden is a 10-foot square holding 100 square feet. An honesty note: the vertex answers "what is the most", but only the setting tells you whether a width of 10 feet fits your yard.

Misconception check. Students often answer a maximum question with only one coordinate, such as "2" for the launch. It is tempting because the vertex formula hands you h first. The cure: a vertex has two parts, and the question usually wants both. "When" is t = 2 seconds; "how high" is h(2) = 144 feet. Say each one in words with its unit.

In practice (your tools). In Desmos, type y = -16x^2 + 64x + 80 and click the top of the curve; Desmos labels it (2, 144). Click where the curve meets the x-axis on the right and read (5, 0), the landing you found in Week 7. For the garden, type y = x(20 - x) and click its top point to read (10, 100).

Pause and reflect. Choose one of these two prompts and answer it in a few sentences.

  • (a) Think of something in your own life that rises and then falls, or falls and then rises, such as the temperature over a day or the crowd at a store. Describe where its turning point is and what the turning point means.
  • (b) Think of a choice where "more" helps only up to a point, such as practice time before a game or water for a houseplant. Describe why the best amount sits in the middle and how you would find it.

Key terms

  • Quadratic function — A function that can be written f(x) = ax² + bx + c with a not equal to zero, so the highest power of the input is 2; its graph is a parabola. (first used: 1. Meet the parabola: a table that turns around)
  • Parabola — The U-shaped curve that is the graph of a quadratic function; it opens either upward or downward and is the same on both sides of a vertical mirror line. (first used: 1. Meet the parabola: a table that turns around)
  • Vertex — The turning point of a parabola, written (h, k): the lowest point when the parabola opens up and the highest point when it opens down. (first used: 2. Find the turning point: the vertex and the axis of symmetry)
  • Axis of symmetry — The vertical line x = h through the vertex of a parabola; points the same horizontal distance from it on either side have equal outputs. (first used: 2. Find the turning point: the vertex and the axis of symmetry)
  • Vertex form — A quadratic function written f(x) = a(x − h)² + k, so the vertex (h, k) and the axis x = h can be read off directly, with h taken from inside the parentheses with its sign reversed. (first used: 3. Read vertex form: the vertex in plain sight)
  • Minimum value — The smallest output a function reaches; for a parabola that opens up (a is positive) it is k, the y-value of the vertex. (first used: 4. Decide up or down: minimum, maximum, and range)
  • Maximum value — The largest output a function reaches; for a parabola that opens down (a is negative) it is k, the y-value of the vertex. (first used: 4. Decide up or down: minimum, maximum, and range)
  • Standard form of a quadratic function — A quadratic function written f(x) = ax² + bx + c with a not equal to zero, so the y-intercept (0, c) can be read off and the vertex can be found from a and b. (first used: 5. Find the vertex from standard form)
  • Vertex formula — The rule h = −b ⁄ (2a) that gives the x-value of a parabola's vertex from standard form; the y-value is then k = f(h). (first used: 5. Find the vertex from standard form)

Summary

  • A quadratic function's graph is a parabola, one smooth U-shaped curve.
  • The vertex is the turning point, and the axis of symmetry is the vertical line through it.
  • In vertex form a(x − h)² + k, the vertex is (h, k); the sign inside the parentheses flips.
  • A positive a opens up to a minimum; a negative a opens down to a maximum.
  • From standard form, find h with −b ⁄ (2a), then evaluate the function at h to get k.
  • The y-intercept is (0, c); the x-intercepts are the solutions of f(x) = 0.
  • The discriminant counts x-intercepts: positive gives two, zero gives one, negative gives none.
  • Expand to reach standard form; complete the square to reach vertex form.
  • A "how high, how much, how big" question is a vertex question; answer both coordinates with units.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. For f(x) = −2(x − 1)² + 8, give the vertex, the axis of symmetry, whether 8 is a minimum or a maximum, the range, and the y-intercept.
  2. For f(x) = x² + 4x − 5, find the vertex with the vertex formula, then predict the number of x-intercepts with the discriminant and find them.
  3. Rewrite 3(x + 2)² − 4 in standard form and check your answer at x = 0.
  4. A bakery prices a box of cookies at p dollars and expects to sell 120 − 20p boxes, so its revenue in dollars is R(p) = p(120 − 20p). Find the price that brings in the most revenue and that revenue.

Answers (worked reasoning — read after attempting)

  1. Match −2(x − 1)² + 8 to a(x − h)² + k: a = −2, h = 1, and k = 8. The vertex is (1, 8), and the axis is x = 1. Since a = −2 is negative, the parabola opens down and 8 is the maximum value. The range is (−∞, 8]. For the y-intercept, set x = 0: −2(0 − 1)² + 8 = −2(1) + 8 = −2 + 8 = 6, so it is (0, 6). Calling 8 a minimum is the trap from Section 4.
  2. Here a = 1, b = 4, and c = −5. The vertex formula gives h = −4 ⁄ (2 · 1) = −4 ⁄ 2 = −2. Then k = f(−2) = (−2)² + 4(−2) − 5 = 4 − 8 − 5 = −9, so the vertex is (−2, −9). The discriminant is 4² − 4(1)(−5) = 16 + 20 = 36, which is positive, so there are two x-intercepts. Factoring, x² + 4x − 5 = (x + 5)(x − 1) = 0, so x = −5 or x = 1. Check: 25 − 20 − 5 = 0 and 1 + 4 − 5 = 0. The axis sits halfway between them: (−5 + 1) ⁄ 2 = −4 ⁄ 2 = −2. It matches.
  3. Square first: (x + 2)² = x² + 4x + 4. Distribute the 3: 3x² + 12x + 12. Combine the constants: 3x² + 12x + 12 − 4 = 3x² + 12x + 8. Check at x = 0: 3(0 + 2)² − 4 = 3(4) − 4 = 12 − 4 = 8, and 3(0)² + 12(0) + 8 = 8. Both forms agree.
  4. Expand: R(p) = 120p − 20p², so a = −20 and b = 120. Since a is negative, the revenue has a maximum. The best price is p = −120 ⁄ (2 · (−20)) = −120 ⁄ (−40) = 3 dollars. At that price the bakery sells 120 − 20(3) = 120 − 60 = 60 boxes, so R(3) = 3 · 60 = 180 dollars. Answer both parts: a price of 3 dollars brings in the most revenue, 180 dollars.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the end-behavior rule reads the graphs of higher-degree polynomials from their leading term.

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