Teach Me Maths

Sketching Graphs and Exponential Graphs

How to sketch linear, quadratic and exponential graphs using intercepts, turning points, growth and decay.

Sketch features before joining points

A sketch should show the correct overall shape and label important features. Find the vertical-axis intercept by setting the input to zero. Find horizontal-axis intercepts by setting the output to zero.

Worked example 1

Find the intercepts and sketch y=3x-6

  1. At x=0, y=-6, giving (0,-6)

  2. Your turn. Set y=0: 3x-6=0. What is x?

    The other intercept is (2,0)

  3. Answer: a straight line through (0,-6) and (2,0), rising from left to right with gradient 3

Notes: parabolas and symmetry

The graph of a quadratic is a smooth U-shaped curve, or an upside-down U, called a parabola. A positive x^2 coefficient gives a U with a minimum; a negative coefficient gives an upside-down U with a maximum.

The axis of symmetry is the vertical line through the turning point: the two sides of the curve are mirror images across it. If there are two distinct roots, it lies halfway between them. Find the turning point and label its coordinates.

Worked example 2

Sketch y=x^2-6x+5, labelling the intercepts and turning point.

  1. Factorise: y=(x-1)(x-5). The horizontal intercepts are (1,0) and (5,0); the vertical intercept is (0,5)

  2. Your turn. The axis of symmetry is halfway between the roots. What is its x value?

    The axis of symmetry is x=3

  3. At x=3, y=-4. Equivalently, y=(x-3)^2-4
  4. Answer: an upward-opening parabola through the three intercepts, with minimum (3,-4)

Worked example 3

Sketch y=-2x^2+8x+10, showing its maximum and intercepts.

  1. Factorise: y=-2(x-5)(x+1). The horizontal intercepts are (5,0) and (-1,0); the vertical intercept is (0,10)
  2. The axis of symmetry is x=2, halfway between the two x-intercepts: x=\frac{-1+5}{2}=2. The maximum lies on this line.

  3. Your turn. To find the maximum height, substitute x=2. Calculate -2(2)^2+8(2)+10

    The maximum is (2,18)

  4. Answer: a downward-opening parabola with maximum (2,18) and the labelled intercepts. In vertex form, y=-2(x-2)^2+18

Recognise growth, decay and an asymptote

An exponential graph has the variable in the power. For y=ab^x with a>0, b>0 and b\ne1, the vertical intercept is (0,a) because b^0=1

If b>1, the graph grows as x increases. If 0<b<1, it decays. Each increase of 1 in x multiplies the output by b

The output stays positive. In one direction the curve gets closer and closer to the line y=0 without reaching it. This line is a horizontal asymptote.

Explore an exponential graph

Try it: compare b=2 and b=\frac12 while keeping a fixed. At b=1 the rule becomes the constant function y=a

Understand negative powers

A negative power means a reciprocal: b^{-x}=(\frac1b)^x. For example, 2^{-3}=\frac{1}{2^3}=\frac18. A negative exponent does not make the output negative.

So ab^{-x}=a(\frac1b)^x. When a>0 and b>1, this is a positive, decreasing curve, like the graphs with a base between zero and one in the explorer.

A negative multiplier is different: if a<0, the graph is reflected in the horizontal axis and its outputs are negative.

Worked example 4

Sketch y=3\times2^x, marking the point where it crosses the y-axis.

  1. The base 2 is greater than 1, and the multiplier 3 is positive, so the curve is positive and increasing. Find its y-intercept by setting x=0.

  2. Your turn. Calculate 3\times2^0.

    2^0=1, so y=3. The curve crosses the y-axis at (0,3).

  3. Answer: a smooth, positive, increasing curve through (0,3), approaching y=0 to the left without touching it. The extra labelled points below are optional guides for drawing the curve.
  4. Optional plotting help: if you want extra points, a short table gives \begin{array}{c|rrrr}x&-1&0&1&2\\y&\frac32&3&6&12\end{array} Each increase of 1 in x doubles the output. You do not need a full table for this sketch.

Worked example 5

Sketch y=12\times2^{-x} and find its value at x=3

  1. Write y=12(\frac12)^x. The base is between 0 and 1, so the curve is positive and decreasing. At x=0, y=12, giving the y-intercept (0,12).

  2. Your turn. At x=3, calculate 12\div2^3

    y=\frac32

  3. Answer: a smooth, positive, decreasing curve through (0,12), approaching y=0 to the right. At x=3, y=\frac32. The extra points (1,6) and (2,3) are optional plotting guides.

Worked example 6

The graph y=ab^{-x} passes through (0,18) and (2,2), with a>0 and b>0. Find a and b.

  1. At x=0, b^0=1, so a=18. At x=2, 2=18b^{-2}=\frac{18}{b^2}

  2. Your turn. Thus b^2=9. Given b>0, find b

    b=3

  3. Answer: a=18, b=3, so y=18\times3^{-x}.

Common mistakes

  • Swapping the intercepts. For an x-intercept set y=0; for a y-intercept set x=0.
  • Guessing a quadratic’s turning point. Find and label the turning point before sketching the curve.
  • Joining exponential points with straight lines. Draw a smooth curve through the points.
  • Making a positive exponential reach its asymptote. For y=ab^x with a,b>0 and b\ne1, the curve approaches y=0 without touching it.
  • Treating a negative exponent as a negative output. A negative exponent means a reciprocal, not a change of sign.

Now try it: Sketching Graphs and Exponential Graphs practice questions

More on this topic: Sketching Graphs and Exponential Graphs worksheet with full solutions