Chapter 8 of 8

All 8 chapters
  1. 1Expanding two brackets, readunderstood
  2. 2Factorising quadratics, readunderstood
  3. 3Using factorising, readunderstood
  4. 4Completing the square, readunderstood
  5. 5The quadratic formula, readunderstood
  6. 6Graphs of quadratics, readunderstood
  7. 7Quadratic sequences, readunderstood
  8. 8A line and a curve, readunderstood

M4.9.8 Chapter 8

A line and a curve

The line y = x + 2 and the curve y = x² cross. How do I find where?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve a linear and a quadratic equation together
  • Solve a quadratic inequality like x² < 9

A point where a line meets a curve is on both. So its x and its y fit both equations. For y = x² and y = x + 2, the two values of y must be equal.

A graph with the x axis x and the y axis y. The curve y = x² and a straight line named y = x + 2 cross at two points, P on the left and Q on the right.PQy = x + 2−3−2−1123134567890xy
The line cuts the curve at P and Q. We find them with algebra, and the graph checks the answer.

Put the two expressions for y equal. That gives x² = x + 2. It is a quadratic equation. Solve it, and each solution is the x of a meeting point.

Draw it first

Where the line meets the curve

  1. A column of equations with arrows between them. The first is x² = x + 2. Step 2 adds an arrow with the note move every term to the left, and x² − x − 2 = 0. Step 3 adds an arrow with the note factorise, and (x − 2)(x + 1) = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = −1.x² = x + 2

    1 Both equations give the same y at a meeting point, so put them equal: x² = x + 2.

  2. A column of equations with arrows between them. The first is x² = x + 2. Step 2 adds an arrow with the note move every term to the left, and x² − x − 2 = 0. Step 3 adds an arrow with the note factorise, and (x − 2)(x + 1) = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = −1.x² = x + 2x² = x + 2x² − x − 2 = 0move every term to the left

    2 Move every term to the left, so that the right side is 0.

  3. A column of equations with arrows between them. The first is x² = x + 2. Step 2 adds an arrow with the note move every term to the left, and x² − x − 2 = 0. Step 3 adds an arrow with the note factorise, and (x − 2)(x + 1) = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = −1.x² = x + 2x² = x + 2x² − x − 2 = 0move every term to the leftx² = x + 2x² − x − 2 = 0(x − 2)(x + 1) = 0move every term to the leftfactorise

    3 Factorise the left side: (x − 2)(x + 1) = 0.

  4. A column of equations with arrows between them. The first is x² = x + 2. Step 2 adds an arrow with the note move every term to the left, and x² − x − 2 = 0. Step 3 adds an arrow with the note factorise, and (x − 2)(x + 1) = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = −1.x² = x + 2x² = x + 2x² − x − 2 = 0move every term to the leftx² = x + 2x² − x − 2 = 0(x − 2)(x + 1) = 0move every term to the leftfactorisex² = x + 2x² − x − 2 = 0(x − 2)(x + 1) = 0x = 2 or x = −1move every term to the leftfactorisesolve each

    4 Solve each bracket. x = 2 or x = −1. These are the x values of the two meeting points.

The x values are only half the answer. A meeting point needs a y as well. Put each x into the line, which is the easier equation, to find it.

Worked example

Find where y = x² and y = x + 2 meet.

  1. At a meeting point both equations give the same y. So x² = x + 2.
  2. Move every term to the left: x² − x − 2 = 0.
  3. Factorise: (x − 2)(x + 1) = 0. So x = 2 or x = −1.
  4. Find y from the line y = x + 2. For x = 2, y = 4. For x = −1, y = 1.
  5. Check in the curve y = x². For x = 2, 2 × 2 = 4. For x = −1, (−1) × (−1) = 1.
  6. The curve and the line meet at (2, 4) and (−1, 1).

Answer: (2, 4) and (−1, 1)

Worked example

Find where y = x² − 2x and y = x + 4 meet.

  1. Put the two values of y equal: x² − 2x = x + 4.
  2. Move every term to the left: x² − 3x − 4 = 0.
  3. Factorise: (x − 4)(x + 1) = 0. So x = 4 or x = −1.
  4. Find y from the line y = x + 4. For x = 4, y = 8. For x = −1, y = 3.
  5. Check in the curve y = x² − 2x. For x = 4, 16 − 8 = 8. For x = −1, 1 + 2 = 3.

Answer: (4, 8) and (−1, 3)

A line does not always cut a curve twice. The quadratic you get can have two solutions, one, or none. This links to the discriminant. First, a line with one solution.

A graph with the x axis x and the y axis y. The curve y = x² and a straight line named y = 2x − 1 touch at one point, P, where the line does not cross the curve.Py = 2x − 1−2−1123−31234567890xy
Here x² = 2x − 1 gives x² − 2x + 1 = 0, which is (x − 1)² = 0. There is one solution, x = 1. The line touches the curve at P.

A line that touches a curve at one point, and does not cross it, is a tangent. Now a line that misses the curve. Take y = x − 3.

A graph with the x axis x and the y axis y. The curve y = x² and a straight line named y = x − 3 never meet. The line stays below the curve.y = x − 3−3−2−1134−7−6−5−2−11234567890xy
Here x² = x − 3 gives x² − x + 3 = 0. Its discriminant is 1 − 12 = −11, which is below 0. There is no solution, so the line misses the curve.

Now a different question. When is x² less than 9? Draw the curve y = x² and the line y = 9. The curve is below the line between the points where they meet.

A graph with the x axis x and the y axis y. The curve y = x² and a dashed line named y = 9 meet at P (−3, 9) and Q (3, 9), with dashed guides to both axes. Between P and Q the curve is below the line.PQy = 9−4−3−2−11234246810120xy
The curve is below the line y = 9 between P and Q, which are at x = −3 and x = 3.

See it three ways

With things

Try the whole numbers from −4 to 4 in x². Mark the ones where x² is less than 9.

As a picture

A number line labelled x, from −5 to 5, with open circles at −3 and 3, a bar between them, and dots at −2, −1, 0, 1 and 2.−5−4−3−2−1012345x

In words

The whole numbers −2, −1, 0, 1 and 2 work. The numbers −3 and 3 give exactly 9, which is not less than 9. So x is between −3 and 3, and not equal to either.

In symbols

x² < 9 means −3 < x < 3.

For x² > 9 it is the other way round. The curve is above the line outside the meeting points. So x is less than −3, or x is more than 3.

A number line labelled x, from −5 to 5, with open circles at −3 and 3, a bar and arrow going left from −3, and a bar and arrow going right from 3.−5−4−3−2−1012345x
x² > 9 means x < −3 or x > 3. The two parts are separate.

Other quadratic inequalities use the same idea. Solve the equation to find the roots. When the x² term is positive, picture the U shaped curve, and read where it is below or above the x axis.

A graph with the x axis x and the y axis y. A curve named y = x² − x − 6 crosses the x axis at two points, A and B, at x = −2 and x = 3. Between A and B the curve is below the x axis.ABy = x² − x − 6−3−2−11234−7−4−3−2−112345670xy
Between A and B the curve is below the x axis, so x² − x − 6 < 0 there.

Worked example

Solve x² − x − 6 < 0.

  1. Solve the equation x² − x − 6 = 0 first: (x − 3)(x + 2) = 0, so x = 3 or x = −2.
  2. These are the roots, where the curve meets the x axis. The curve is a U shape, so it is below the axis between the roots.
  3. So x² − x − 6 is below 0 when x is between −2 and 3: −2 < x < 3.
  4. Check x = 0, which is in the range. Then x² − x − 6 = −6, which is below zero.
  5. Check x = 4, which is outside the range: 16 − 4 − 6 = 6.

Answer: −2 < x < 3

Try it

Test numbers in an inequality

You need: paper and a pencil.

  1. Take x² < 9. Try x = −4, −2, 0, 2 and 4.
  2. For each one work out x², and write yes if it is less than 9 and no if it is not.
  3. Look at which numbers say yes. Are they in one run?

What you will see: The yes answers are −2, 0 and 2. They are in one run, between −3 and 3. The no answers are on the outside.

If you cannot try it, this is what happens.
A table with the heads x, x² and x² < 9. Rows: −4 gives 16 and no; −2 gives 4 and yes; 0 gives 0 and yes; 2 gives 4 and yes; 4 gives 16 and no.xx²x² < 9−416no−24yes00yes24yes416no

A wrong idea many people have

x² < 9 means x < 3.

Why it feels right: The square root of 9 is 3, so it seems that x must be below 3.

The test that breaks it: Put x = −5. Then x < 3 is true, but x² = 25, which is not less than 9.

What is true: x can be negative too. The numbers between −3 and 3 all work, so x² < 9 means −3 < x < 3.

A wrong idea many people have

x² > 9 means x > 3.

Why it feels right: It is the same idea as before, with the sign turned round, so only the number 3 seems to matter.

The test that breaks it: Put x = −4. Then x² = 16, which is more than 9. But −4 is not more than 3.

What is true: Large negative numbers also have large squares. So x² > 9 means x < −3 or x > 3.

A wrong idea many people have

When a line meets a curve at x = −1, the meeting point is (−1, 0).

Why it feels right: Solving the quadratic gives only x values, and writing 0 for y seems to complete the point.

The test that breaks it: For y = x + 2 and x = −1, the line gives y = 1. The point (−1, 0) is not on the line.

What is true: Find y from the line, the easier equation. Give each meeting point as a pair, (x, y). Here it is (−1, 1).

Key words

tangent
a line that touches a curve at one point and does not cross it there.

Back to your guess

The answer is A: Find the x where x² = x + 2, then find y from the line.

At a point where they cross, both equations give the same y. So x² = x + 2. That is a quadratic, which you can solve. Then put each x into the line to find y. The crossing points are (−1, 1) and (2, 4).

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The line y = 2x + 3 meets the curve y = x². Which points are they?

    Tap your answer.

    1. Yes, that is right. x² = 2x + 3 gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1. The line gives y = 9 and y = 1.

    2. Not this one. Put x = −1 into the line y = 2x + 3. It gives 2 × (−1) + 3 = 1, not −5. Go to M4.9.8. Try another.

    3. Not this one. You used y = 2x and left out the + 3. Put each x into y = 2x + 3. Go to M4.9.8. Try another.

    4. Not this one. These are only the x values. A meeting point needs a y as well, from the line. Go to M4.9.8. Try another.

    See the answer

    (3, 9) and (−1, 1)

    x² = 2x + 3 gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1. The line gives y = 9 and y = 1.

    • If you chose B: Put x = −1 into the line y = 2x + 3. It gives 2 × (−1) + 3 = 1, not −5. Go to M4.9.8.
    • If you chose C: You used y = 2x and left out the + 3. Put each x into y = 2x + 3. Go to M4.9.8.
    • If you chose D: These are only the x values. A meeting point needs a y as well, from the line. Go to M4.9.8.

  2. Which equation do you get from y = x² and y = x + 6?

    Tap your answer.

    1. Not this one. Taking x + 6 from both sides changes the signs of both terms, giving x² − x − 6 = 0. Go to M4.9.8. Try another.

    2. Not this one. The 6 changes sign too when it moves to the left. It becomes −6. Go to M4.9.8. Try another.

    3. Yes, that is right. The two values of y are equal: x² = x + 6. Take x + 6 from both sides: x² − x − 6 = 0.

    4. Not this one. You have lost the x² term. Put the two values of y equal: x² = x + 6. Go to M4.9.8. Try another.

    See the answer

    x² − x − 6 = 0

    The two values of y are equal: x² = x + 6. Take x + 6 from both sides: x² − x − 6 = 0.

    • If you chose A: Taking x + 6 from both sides changes the signs of both terms, giving x² − x − 6 = 0. Go to M4.9.8.
    • If you chose B: The 6 changes sign too when it moves to the left. It becomes −6. Go to M4.9.8.
    • If you chose D: You have lost the x² term. Put the two values of y equal: x² = x + 6. Go to M4.9.8.

  3. The line y = 4x − 4 meets the curve y = x². What happens?

    Tap your answer.

    1. Not this one. The equation x² = 4x − 4 gives x² − 4x + 4 = 0, which is (x − 2)² = 0. That has only one solution. Go to M4.9.8. Try another.

    2. Not this one. Try x = 2. The curve gives 4 and the line gives 4 × 2 − 4 = 4. They meet there. Go to M4.9.8. Try another.

    3. Not this one. Put x = 4 into the line: 4 × 4 − 4 = 12, not 16. The meeting point has x = 2. Go to M4.9.8. Try another.

    4. Yes, that is right. x² = 4x − 4 gives x² − 4x + 4 = 0, so (x − 2)² = 0 and x = 2. The line gives y = 4 × 2 − 4 = 4. It touches the curve at (2, 4).

    See the answer

    They meet at one point, (2, 4), where the line touches the curve.

    x² = 4x − 4 gives x² − 4x + 4 = 0, so (x − 2)² = 0 and x = 2. The line gives y = 4 × 2 − 4 = 4. It touches the curve at (2, 4).

    • If you chose A: The equation x² = 4x − 4 gives x² − 4x + 4 = 0, which is (x − 2)² = 0. That has only one solution. Go to M4.9.8.
    • If you chose B: Try x = 2. The curve gives 4 and the line gives 4 × 2 − 4 = 4. They meet there. Go to M4.9.8.
    • If you chose C: Put x = 4 into the line: 4 × 4 − 4 = 12, not 16. The meeting point has x = 2. Go to M4.9.8.

  4. Solve x² < 16.

    Tap your answer.

    1. Not this one. Put x = −5. Then x < 4 is true, but x² = 25, which is not below 16. The negative numbers matter too. Go to M4.9.8. Try another.

    2. Yes, that is right. The curve y = x² is below the line y = 16 between x = −4 and x = 4. So −4 < x < 4.

    3. Not this one. This is the answer to x² > 16, where the square is large. For x² < 16 the square is small, so x is near 0. Go to M4.9.8. Try another.

    See the answer

    −4 < x < 4

    The curve y = x² is below the line y = 16 between x = −4 and x = 4. So −4 < x < 4.

    • If you chose A: Put x = −5. Then x < 4 is true, but x² = 25, which is not below 16. The negative numbers matter too. Go to M4.9.8.
    • If you chose C: This is the answer to x² > 16, where the square is large. For x² < 16 the square is small, so x is near 0. Go to M4.9.8.

  5. Solve x² > 25.

    Tap your answer.

    1. Not this one. Put x = −6. Then x² = 36, which is more than 25, but −6 is not more than 5. Go to M4.9.8. Try another.

    2. Not this one. This is the answer to x² < 25. For x² > 25 the square is large, so x is far from 0. Go to M4.9.8. Try another.

    3. Yes, that is right. The curve y = x² is above the line y = 25 outside x = −5 and x = 5. So x < −5 or x > 5.

    See the answer

    x < −5 or x > 5

    The curve y = x² is above the line y = 25 outside x = −5 and x = 5. So x < −5 or x > 5.

    • If you chose A: Put x = −6. Then x² = 36, which is more than 25, but −6 is not more than 5. Go to M4.9.8.
    • If you chose B: This is the answer to x² < 25. For x² > 25 the square is large, so x is far from 0. Go to M4.9.8.

  6. Solve x² − 2x − 8 < 0.

    Tap your answer.

    1. Yes, that is right. x² − 2x − 8 = (x − 4)(x + 2), so the roots are −2 and 4. The U shaped curve is below the x axis between them: −2 < x < 4.

    2. Not this one. This is where the curve is above the x axis. The inequality asks where it is below, which is between the roots. Go to M4.9.8. Try another.

    3. Not this one. The brackets are (x − 4)(x + 2). Putting x − 4 = 0 gives x = 4, and x + 2 = 0 gives x = −2. Go to M4.9.3. Try another.

    See the answer

    −2 < x < 4

    x² − 2x − 8 = (x − 4)(x + 2), so the roots are −2 and 4. The U shaped curve is below the x axis between them: −2 < x < 4.

    • If you chose B: This is where the curve is above the x axis. The inequality asks where it is below, which is between the roots. Go to M4.9.8.
    • If you chose C: The brackets are (x − 4)(x + 2). Putting x − 4 = 0 gives x = 4, and x + 2 = 0 gives x = −2. Go to M4.9.3.

  7. From an earlier chapter

    Factorise x² − 9x + 14.

    Tap your answer.

    1. Not this one. These multiply to 14 and add to +9. The x number is −9, so both numbers are negative. Go to M4.9.2. Try another.

    2. Not this one. These multiply to 14, but they add to −15, not −9. Go to M4.9.2. Try another.

    3. Not this one. These multiply to −14. The last number is +14, so both numbers have the same sign. Go to M4.9.2. Try another.

    4. Yes, that is right. Find two numbers with product 14 and sum −9. The pair −2 and −7 works: (−2) × (−7) = 14 and (−2) + (−7) = −9.

    See the answer

    (x − 2)(x − 7)

    Find two numbers with product 14 and sum −9. The pair −2 and −7 works: (−2) × (−7) = 14 and (−2) + (−7) = −9.

    • If you chose A: These multiply to 14 and add to +9. The x number is −9, so both numbers are negative. Go to M4.9.2.
    • If you chose B: These multiply to 14, but they add to −15, not −9. Go to M4.9.2.
    • If you chose C: These multiply to −14. The last number is +14, so both numbers have the same sign. Go to M4.9.2.

  8. The line y = 7x − 10 meets the curve y = x² at two points. Give the two x values, smaller first.

    See the answer

    x = 2 and x = 5

    x² = 7x − 10 gives x² − 7x + 10 = 0, so (x − 2)(x − 5) = 0. The x values are 2 and 5.

  9. How many whole numbers x, negative, zero or positive, satisfy x² < 20?

    See the answer

    9

    x² < 20 means x is between about −4.47 and 4.47. The whole numbers are −4, −3, −2, −1, 0, 1, 2, 3 and 4. That is 9 numbers.

Keep it

Copy this into your exercise book.

Put the two y values equal, solve, then find y from the line. For x² < 9, x is between −3 and 3. For x² > 9, x is below −3 or above 3.

Draw: A U shaped curve with a flat line across it, and the two meeting points marked and named.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (898 KB). It opens on any phone.