Algebra, Year 11: trigonometric graphs, transformations and tangents

360

HintOne full turn.

The whyThe sine wave starts at 0, rises to 1 at 90°, returns to 0 at 180°, falls to −1 at 270° and is back to 0 at 360°, then does the same again. The cosine graph also repeats every 360°.

180

HintIt is half the figure for the sine and cosine graphs.

The whyThe tangent graph is a series of separate, identical branches, each rising steeply from far below the x-axis to far above it. One branch passes through 0°, the next through 180°, and so on, so tan x = tan (x + 180°).

−1 and 1

HintThink of the largest value a sine can have in a right-angled triangle.

The whyBoth waves go no higher than 1 and no lower than −1. So an equation such as sin x = 1.5 has no solutions. The tangent graph is different: it takes every value.

tan 90° has no value: as x approaches 90° the curve climbs without limit

HintThink of a right-angled triangle whose angle is getting closer and closer to a right angle.

The whytan is opposite ÷ adjacent. As the angle nears 90°, the adjacent side shrinks towards nothing, so the ratio becomes enormous. The graph shoots upwards just before 90° and reappears from far below just after it, with the same break at 270°.

300

HintBetween 0° and 360° the cosine curve is a mirror image of itself in the line x = 180°.

The whyThe cosine graph from 0° to 360° is symmetrical about x = 180°. 60° is 60° after the start, so the matching angle is 60° before the end: 360° − 60° = 300°.

A translation of 3 units up, parallel to the y-axis

HintEvery y-value on the curve is increased by the same amount.

The whyAdding 3 outside the function adds 3 to every output, so each point moves 3 up and the shape is unchanged. As a vector the translation is 0 across and 3 up. y = f(x) − 3 would move the graph 3 down.

A translation of 2 units to the left, parallel to the x-axis

HintAsk what x must be to get the same output as before.

The whyTo get the output that f used to give at x = 5, the new function needs x = 3, because 3 + 2 = 5. Every point is reached 2 earlier, so the graph moves 2 to the left. A change inside the bracket affects x and works the 'opposite' way to what the sign suggests.

A reflection of the graph of y = f(x) in the x-axis

HintEvery point keeps its x-value, but its height changes sign.

The whyMultiplying the output by −1 turns every y-value into its opposite, so a point 4 above the x-axis moves to 4 below it. Points on the x-axis stay where they are. Changing to y = f(−x) reflects in the y-axis instead.

(−2, 5)

HintThe new function needs the opposite input to give the same output.

The whyf(−x) gives the output 5 when −x = 2, that is when x = −2. Every point keeps its height and swaps sides, so the graph is reflected in the y-axis.

y = (x − 4)²

HintThe new curve must have its lowest point where x is 4.

The whyA translation of 4 to the right replaces x by (x − 4), giving y = (x − 4)². Check: the turning point should now be at (4, 0), and (4 − 4)² = 0.

perpendicular

HintThe two lines meet at 90°.

The whyThis circle theorem is what makes the algebra work: find the gradient of the radius, and the tangent's gradient is its negative reciprocal. For a circle centred on the origin, the radius to the point (a, b) has gradient b/a.

3; −1/3; 20/3

HintWork in three steps: radius, then perpendicular, then use the point.

The whyGradient of radius = 6 ÷ 2 = 3. The tangent is perpendicular, so its gradient is −1/3. Using y = −(1/3)x + c at (2, 6): 6 = −2/3 + c, so c = 20/3. The tangent is y = −(1/3)x + 20/3, or x + 3y = 20.

3x + 4y = 25

HintStart with the gradient of the line from the centre to the point.

The whyThe radius from (0, 0) to (3, 4) has gradient 4/3, so the tangent has gradient −3/4. Then y − 4 = −(3/4)(x − 3), which gives y = −(3/4)x + 25/4. Multiplying through by 4 gives 3x + 4y = 25.

y = 5

HintPicture the very top of the circle.

The why(0, 5) is the highest point of the circle, and the radius to it is vertical. The tangent is at right angles to the radius, so it is horizontal: y = 5. The negative-reciprocal rule cannot be used here because a vertical line has no gradient.

5

HintThe shortest route from the centre to a tangent runs along a radius.

The whyThe radius to the point of contact meets the tangent at right angles, and the perpendicular distance is the shortest distance from a point to a line. So the distance equals the radius, √25 = 5.