Recognising and Sketching Graph Shapes

Worked example

1. 1

   $y=3-x^2$ — The x² has a minus, so the U turns upside down. Its top is at (0, 3), because 3 minus a square can never be more than 3.

2. 2

   $y=x^3+1$ — The x³ is positive: a big negative x gives a big negative y. So the curve is low on the left and high on the right.

3. 3

   $x=0\Rightarrow y=0+1=1$ — Put x = 0 to find where it crosses the y-axis: at (0, 1).

4. 4

   $y=2+x-4x^2-x^3$ — Find the highest power first: −x³. It is negative, so the curve is high on the left and low on the right.

5. 5

   $x=0\Rightarrow y=2$ — Every x term becomes 0, leaving the number on its own. So it crosses the y-axis at (0, 2).

Worked example

1. 1

   $y=\frac{1}{x}$ — Positive x gives positive y, and negative x gives negative y. So the parts sit top right and bottom left.

2. 2

   $y=\frac{-4}{x}$ — The minus flips the sign of every y. So the parts move to top left and bottom right.

3. 3

   $y=\frac{6}{x^2}$ — x² is never negative, so y is always positive. Both parts sit above the x-axis.

4. 4

   $\frac{6}{x^2}\ne 0$ — 6 divided by any number is never 0. So the curve never touches the x-axis either.

Worked example

1. 1

   $y=(x+2)(x-1)(x-4)$ — The question. The three x's multiply to +x³, so the curve is low on the left and high on the right.

2. 2

   $(x+2)(x-1)(x-4)=0$ — The curve meets the x-axis where y = 0. Three brackets multiply to 0 only when one of them is 0.

3. 3

   $x=-2,x=1,x=4$ — Set each bracket to 0. x + 2 = 0 gives x = −2, and the other two give 1 and 4.

4. 4

   $x=0:(0+2)(0-1)(0-4)$ — Now put x = 0 to find where the curve meets the y-axis.

5. 5

   $=2\times (-1)\times (-4)=8$ — Two negatives multiply to a positive. So it crosses the y-axis at (0, 8).

6. 6

   $\text{up through }-2,\text{down through }1,\text{up through }4$ — Join them: start low, rise through −2, turn, fall through 8 on the y-axis and through 1, turn, rise through 4. Write −2, 1, 4 and 8 on the sketch.

Worked example

1. 1

   $y=x(x-3{)}^2$ — The question. x times x² gives +x³, so the curve is low on the left and high on the right.

2. 2

   $x=0\text{or}(x-3{)}^2=0$ — y = 0 when one of the factors is 0.

3. 3

   $x=0,x=3$ — Two x-values. x = 3 comes from a squared bracket, so it is the repeated root.

4. 4

   $x=0:y=0\times (-3{)}^2=0$ — Put x = 0 for the y-axis. The curve goes through the origin, (0, 0).

5. 5

   $(x-3{)}^2\ge 0$ — Near x = 3, x is positive and a square is never negative. So y is never below 0 there: the curve comes down, touches (3, 0) and goes back up.

6. 6

   $\text{up through }0,\text{turn},\text{touch at }3$ — The sketch: rise through the origin, turn at a top between 0 and 3, come down to touch the axis at 3, then rise again.

Worked example

1. 1

   $y=x^3-4x+1$ — The curve. Fill its table one x at a time.

2. 2

   $x=-2:(-2{)}^3-4(-2)+1$ — Put −2 in brackets everywhere x appears, so no minus sign gets lost.

3. 3

   $(-2{)}^3=-8$ — Three negatives multiplied together give a negative.

4. 4

   $-4\times (-2)=+8$ — Minus times minus is plus.

5. 5

   $-8+8+1=1$ — Add the three terms. So y = 1 when x = −2.

6. 6

   $x=1.5:3.375-6+1=-1.625$ — Work out each term: 1.5³ = 3.375 and 4 × 1.5 = 6. Then add.

7. 7

   $-1.625\approx -1.6$ — Round to the same number of decimal places as the other values in the table.

Worked example

1. 1

   $(-2,1),(-1,4),(0,1),(1,-2),(2,1)$ — Plot each pair from the table: x across, then y up.

2. 2

   $y=x^3-4x+1$ — Join the points freehand with one smooth curve. It must pass through every point.

3. 3

   $x^3-4x+1=1$ — To solve this, notice the left side is the curve's y. So draw the line y = 1.

4. 4

   $x=-2,x=0,x=2$ — Read the x-value at each place the line meets the curve. Three crossings, so three solutions.

5. 5

   $2^3-4(2)+1=8-8+1=1$ — Check one answer in the equation. It gives 1, so x = 2 is right.

6. 6

   $\text{lowest point}\approx (1.2,-2.1)$ — A question may ask for a turning point. Read its x and y off your graph, to the nearest small square.