HA12 - Exponential and trigonometric graphs | Eduqas GCSE Mathematics

Recognise and sketch exponential, sine, cosine and tangent graphs, including negative inputs and angles beyond one turn. HA12 uses degrees for trigonometric arguments throughout.

Exponential graphs

A sequence of heights doubles every step. Should its graph be a straight line? Equal input steps multiply the output by a fixed factor, rather than adding a fixed amount, so an exponential graph bends.

An exponential function has the input in the power: $y=k^x$ , where the base $k$ is positive. For example, $2^{-2}=1/2^2=1/4$ , and $2^{1/2}=\sqrt2$ . Tables of integer inputs give useful anchors; join them with the smooth exponential shape, rather than straight segments.

Exponential graph features

Every $y=k^x$ , with $k>0$ , passes through $(0,1)$ and has positive outputs for all real $x$ .

If $k>1$ , the graph increases: exponential growth.
If $0<k<1$ , it decreases: exponential decay.
If $k=1$ , it is the constant graph $y=1$ .
For the two non-constant cases, $y=0$ is a horizontal asymptote: outputs approach zero in one direction but never equal it. There is no $x$ -intercept.

The base must be positive for this whole-real-input graph. For example, a negative base would not have a real value at every fractional input. The excluded base $0$ also fails at zero and negative inputs.

Worked example 1: exponential growth

Question. Sketch $y=2^x$ for $-2\le x\le3$ , showing its intercept and behaviour towards the horizontal axis.

Plan. Calculate anchors, locate $(0,1)$ , and use the increasing exponential shape.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$2^x$	$1/4$	$1/2$	$1$	$2$	$4$	$8$

Draw a smooth increasing curve through these points. It crosses the vertical axis at $1$ , remains above the horizontal axis and approaches that axis as $x$ becomes very negative.

Answer. An increasing curve through the six anchors, with $y$ -intercept $1$ and asymptote $y=0$ .

Check. Successive outputs have ratio $2$ , while their differences are not constant. No anchor has a zero or negative output, consistent with the required shape.

Change only the base from $2$ to its reciprocal $1/2$ . Every step now halves the output.

Worked example 2: exponential decay

Question. Sketch $y=(1/2)^x$ over the same interval, showing its intercept and asymptotic behaviour.

Plan. Use the same input anchors and compare the outputs in order.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$(1/2)^x$	$4$	$2$	$1$	$1/2$	$1/4$	$1/8$

The curve decreases smoothly through $(0,1)$ and approaches $y=0$ as $x$ increases. It does not reach or cross the axis.

Answer. A decreasing positive curve through the anchors, with $y$ -intercept $1$ and asymptote $y=0$ .

Check. Every successive output has ratio $1/2$ . Also $(1/2)^x=2^{-x}$ , so matching opposite inputs in the first table gives these outputs.

Both graphs stay positive even when their inputs are negative.

[DIAGRAM: asset_name: Exponential growth and decay; asset_slug: exponential-pair; caption: Changing the positive base across one changes growth to decay.; description: Exact graphs y=2 to x and y=(1/2) to x use labelled axes and integer anchors over x=-2 to3. Both cross(0,1), remain positive and approach the horizontal axis in opposite directions. A note identifies the separate constant-base-one case.]
Changing the positive base across one changes growth to decay.

Watch out

$2^x$ and $2x$ are different functions. At $x=3$ , they give $8$ and $6$ . A negative input also does not make an exponential output negative: $2^{-3}=1/8$ .

Use a fresh base and distinguish growth from decay.

Sine and cosine

To extend acute-angle trigonometry, rotate a radius of length $1$ from the positive horizontal axis. Its endpoint is $(\cos x,\sin x)$ . This agrees with adjacent/hypotenuse and opposite/hypotenuse in the first quadrant and defines the signed coordinates in all four quadrants.

Anticlockwise angles are positive; clockwise angles are negative. A full turn returns to the same point, so adding or subtracting $360^\circ$ does not change sine or cosine. For example, $450^\circ=360^\circ+90^\circ$ has endpoint $(0,1)$ .

The circle coordinates explain both the signs and the repeated waves.

[DIAGRAM: asset_name: Sine and cosine around a full turn; asset_slug: full-turn-coordinates; caption: Cosine is horizontal position and sine is vertical position on the unit circle.; description: Exact unit circle shows axis endpoints at0,90,180,270 degrees and a150-degree endpoint(-sqrt3/2,1/2) with projections. Quadrant signs of sine and cosine are labelled outside the circle.]

A graph's period is the smallest positive input shift that repeats the whole pattern. The amplitude of a sine or cosine wave is the distance from its midline to a maximum. For $y=\sin x$ and $y=\cos x$ , period is $360^\circ$ , midline $y=0$ , amplitude $1$ , and output range $-1\le y\le1$ .

$x$ in degrees	$0$	$90$	$180$	$270$	$360$
$\sin x$	$0$	$1$	$0$	$-1$	$0$
$\cos x$	$1$	$0$	$-1$	$0$	$1$

Draw smooth waves through the anchors, repeating in both directions. Sine starts at zero and rises; cosine starts at its maximum. To solve a trigonometric equation graphically, draw the horizontal output level and read every intersection in the stated interval, including valid endpoints.

Worked example 3: reading the sine wave

Question. Sketch $y=\sin x$ over $-180^\circ\le x\le540^\circ$ . State $\sin450^\circ$ and solve $\sin x=1/2$ for $0^\circ\le x\le360^\circ$ .

Plan. Repeat the quadrant anchors, then use the two intersections with height $1/2$ .

$x$	$-180^\circ$	$-90^\circ$	$0^\circ$	$90^\circ$	$180^\circ$	$270^\circ$	$360^\circ$	$450^\circ$	$540^\circ$
$\sin x$	$0$	$-1$	$0$	$1$	$0$	$-1$	$0$	$1$	$0$

The level $y=1/2$ meets the wave at $30^\circ$ and $150^\circ$ during the first full turn. These correspond to unit-circle points above the axis with the same vertical coordinate.

Answer. The smooth wave follows the anchors; $\sin450^\circ=1$ . The equation solutions are $x=30^\circ,150^\circ$ .

Check. $450^\circ$ is coterminal with $90^\circ$ , giving vertical coordinate $1$ . Reflection of the $30^\circ$ radius across the vertical axis gives $150^\circ$ , preserving its vertical coordinate $1/2$ .

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HA12 - Exponential and trigonometric graphs

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Sine and cosine

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