Build an equation from a graph

A graph has maximum 8, minimum −2, period 10 and an upward midline crossing at x = 2. Write a positive sine equation. Upward means the height passes through the middle from below to above.

1. A = 8 − (−2)2$\frac{8-(-2)}{2}$ = 5 and B = 8 + (−2)2$\frac{8+(-2)}{2}$ = 3.Half the span gives the amplitude; average gives the center.
2. ω = 2π10$\frac{2\pi }{10}$ = π5$\frac{\pi }{5}$.The inside must advance one full turn during 10 horizontal units.
3. Choose positive sine with h = 2, the upward middle crossing.Sine begins at 0 and rises, so after adding 3 it starts at the center and rises.
4. φ = ωh = π5$\frac{\pi }{5}$ × 2 = 2π5$\frac{2\pi }{5}$.Convert horizontal position into inside offset.
5. Plug back: π5$\frac{\pi }{5}$ × 2 − 2π5$\frac{2\pi }{5}$ = 0.The crossing is at the intended inside start.
6. Write y = 5 sin(π5$\frac{\pi }{5}$x − 2π5$\frac{2\pi }{5}$) + 3.The positive A preserves the upward direction.

A smooth sinusoidal graph has maximum 9, minimum 3 and consecutive peaks at x = 1 and x = 9. Write a positive cosine equation. This asks you to turn its heights, width and peak position into a formula.

1. A = 9 − 32$\frac{9-3}{2}$ = 3 and B = 9 + 32$\frac{9+3}{2}$ = 6.The two extreme heights determine size and center.
2. T = 9 − 1 = 8; ω = 2π8$\frac{2\pi }{8}$ = π4$\frac{\pi }{4}$.Adjacent peaks are the same stage one cycle apart.
3. Choose cosine with positive A and h = 1.At the chosen start, positive cosine sits at a peak.
4. φ = ωh = π4$\frac{\pi }{4}$ × 1 = π4$\frac{\pi }{4}$.φ is an inside offset, so multiply the x-distance by the inside coefficient.
5. Plug h back: π4$\frac{\pi }{4}$ × 1 − π4$\frac{\pi }{4}$ = 0.The chosen peak position feeds cosine its starting input.
6. Write y = 3 cos(π4$\frac{\pi }{4}$x − π4$\frac{\pi }{4}$) + 6, or y = 3 cos(π4$\frac{\pi }{4}$(x − 1)) + 6.Factoring displays the horizontal distance without changing the function.

Use the same graph with peaks at x = 1 and 9, maximum 9 and minimum 3. Write a positive sine equation instead of cosine. This asks for a sine start that produces the same highs and lows as the cosine picture.

1. A = 3, B = 6, T = 8, ω = π4$\frac{\pi }{4}$.Changing the wave name does not change the graph's size, center or period.
2. An upward middle crossing occurs one quarter period before the peak at 1: h = 1 − 2 = −1.Positive sine rises from its middle to its peak during the first quarter cycle.
3. φ = ωh = −π4$\frac{\pi }{4}$.The selected sine start lies one unit left of 0.
4. Plug back: π4$\frac{\pi }{4}$(−1) + π4$\frac{\pi }{4}$ = 0.That crossing is the inside sine start.
5. Write y = 3 sin(π4$\frac{\pi }{4}$x + π4$\frac{\pi }{4}$) + 6.A negative φ produces a plus in the standard sine input.

For y = 3 cos(π4$\frac{\pi }{4}$(x − 1)) + 6, locate the five Key points of one cycle. This asks for the peak, middle, trough, middle and next peak.

1. The start is h = 1 and T = 8, so Quarter period = T ÷ 4 = 2.A full turn has four quarter-turn steps.
2. Use x = h + k(T4$\frac{T}{4}$) with k = 0, 1, 2, 3, 4; the table shows these inputs.Each step advances the inside by (π4$\frac{\pi }{4}$) × 2 = π2$\frac{\pi }{2}$.
3. In the table, the column under 1 has inside 0, so y = 3(1) + 6 = 9.Positive cosine starts at a peak.
4. The columns under 3 and 7 have inside π2$\frac{\pi }{2}$ and 3π2$\frac{3\pi }{2}$, so y = 6.Cosine gives 0 at the two middle stages.
5. The column under 5 has inside π, so y = 3(−1) + 6 = 3; the column under 9 returns to inside 2π and output 9.The half-turn gives the trough and the full turn gives the next peak.