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Find the amplitude, period, phase shift and midline of y = 4 sin(π4$\frac{\pi }{4}$x + π2$\frac{\pi }{2}$) − 2. Say which way the graph is shifted, then check the shift and the range.

1. Look at the x-coefficient inside the sine. In π4$\frac{\pi }{4}$x + π2$\frac{\pi }{2}$ it is +π4$\frac{\pi }{4}$, which is positive, so there is no −1 to factor out. The inside stays π4$\frac{\pi }{4}$x + π2$\frac{\pi }{2}$.A −1 is factored out only when the number multiplying x is negative. A positive coefficient already gives ω > 0, which the period and phase-shift formulas need.
2. Use neither rule. The function stays y = 4 sin(π4$\frac{\pi }{4}$x + π2$\frac{\pi }{2}$) − 2, and the 4 in front keeps its sign.The sine odd rule sin(−u) = −sin u is used only after a −1 has been pulled out of the inside. Nothing was pulled out, so A is not flipped to −4.
3. Write the inside in the form ωx − φ: π4$\frac{\pi }{4}$x + π2$\frac{\pi }{2}$ = π4$\frac{\pi }{4}$x − (−π2$\frac{\pi }{2}$). Match y = A sin(ωx − φ) + B to get A = 4, ω = π4$\frac{\pi }{4}$, φ = −π2$\frac{\pi }{2}$ and B = −2.The form subtracts φ, so adding π2$\frac{\pi }{2}$ means φ = −π2$\frac{\pi }{2}$. The − 2 outside means B = −2, not 2. Keeping these signs makes the shift and the midline come out right.
4. Amplitude = |A| = |4| = 4.Amplitude is the distance from the midline to a peak. It is the size of A and is never negative.
5. Period = 2πω$\frac{2\pi }{\omega }$ = 2π ÷ π4$\frac{\pi }{4}$ = 2π × 4π$\frac{4}{\pi }$ = 8.Sine repeats each time its inside grows by 2π, and the inside grows by π4$\frac{\pi }{4}$ for every 1 unit of x.
6. Phase shift = φω$\frac{\phi }{\omega }$ = −π2$\frac{\pi }{2}$ ÷ π4$\frac{\pi }{4}$ = −π2$\frac{\pi }{2}$ × 4π$\frac{4}{\pi }$ = −2. The graph is shifted 2 units to the left.A sine cycle starts where its inside is 0, which happens at x = φω$\frac{\phi }{\omega }$. A negative shift puts that start to the left of x = 0.
7. Midline: y = B = −2.Adding B moves the whole curve up or down, so the curve is centred on the line y = B.
8. Plug the shift x = −2 into the inside. Steps 1 and 2 left it unchanged, so the original inside and the rewritten inside are the same: π4$\frac{\pi }{4}$(−2) + π2$\frac{\pi }{2}$ = −π2$\frac{\pi }{2}$ + π2$\frac{\pi }{2}$ = 0. The range runs from B − |A| = −2 − 4 = −6 to B + |A| = −2 + 4 = 2, so it is [−6, 2].The inside must be 0 at the phase shift, because that is where the cycle starts on the midline. The curve reaches exactly |A| above and below the midline, so the range is [B − |A|, B + |A|].

Find the amplitude, period, phase shift and midline of y = 3 cos(−π6$\frac{\pi }{6}$x + 2π3$\frac{2\pi }{3}$) − 1. Say which way the graph is shifted. Then check the shift in both the original and the rewritten inside, and check the range. The figure is the finished check sketch. It shows the midline and the five key points of one cycle.

1. The x-coefficient inside is −π6$\frac{\pi }{6}$, which is negative. Factor −1 out of the whole inside: −π6$\frac{\pi }{6}$x + 2π3$\frac{2\pi }{3}$ = −(π6$\frac{\pi }{6}$x − 2π3$\frac{2\pi }{3}$).The −1 has to come out of both terms, so +2π3$\frac{2\pi }{3}$ turns into −2π3$\frac{2\pi }{3}$ inside the bracket. Multiplying back out gives −π6$\frac{\pi }{6}$x + 2π3$\frac{2\pi }{3}$ again, so only the look has changed.
2. Use the cosine even rule cos(−u) = cos u with u = π6$\frac{\pi }{6}$x − 2π3$\frac{2\pi }{3}$: y = 3 cos(−(π6$\frac{\pi }{6}$x − 2π3$\frac{2\pi }{3}$)) − 1 = 3 cos(π6$\frac{\pi }{6}$x − 2π3$\frac{2\pi }{3}$) − 1.Cosine is even, so the −1 in front of u drops away and the 3 out front stays 3. Only the sine odd rule would move a minus sign onto A.
3. Match y = 3 cos(π6$\frac{\pi }{6}$x − 2π3$\frac{2\pi }{3}$) − 1 to y = A cos(ωx − φ) + B: A = 3, ω = π6$\frac{\pi }{6}$, φ = 2π3$\frac{2\pi }{3}$, B = −1.The form has a minus in front of φ, so − 2π3$\frac{2\pi }{3}$ means φ = +2π3$\frac{2\pi }{3}$. The − 1 at the end means B = −1. Each sign is kept exactly as the form reads it.
4. Amplitude = |A| = |3| = 3.The amplitude is the height of the peaks above the midline. It is the size of A, so it is never negative.
5. Period = 2πω$\frac{2\pi }{\omega }$ = 2π ÷ π6$\frac{\pi }{6}$ = 2π · 6π$\frac{6}{\pi }$ = 12.One cycle of cosine uses 2π of inside. With the speed now positive, the x-length of a cycle is 2π divided by ω. Dividing by the original −π6$\frac{\pi }{6}$ would have given −12, a negative length, which is why the −1 was factored out first.
6. Phase shift = φω$\frac{\phi }{\omega }$ = 2π3$\frac{2\pi }{3}$ ÷ π6$\frac{\pi }{6}$ = 2π3$\frac{2\pi }{3}$ · 6π$\frac{6}{\pi }$ = 4, so the graph is shifted 4 units to the right.A positive φω$\frac{\phi }{\omega }$ means a shift to the right. The plus sign in the original inside does not mean left. Once −1 is factored out it becomes the minus in ωx − φ, and that sign sets the direction.
7. Midline: y = B = −1.B is the vertical shift, so the graph swings 3 units above and 3 units below the line y = −1.
8. Plug the shift x = 4 into both insides. Original: −π6$\frac{\pi }{6}$(4) + 2π3$\frac{2\pi }{3}$ = −2π3$\frac{2\pi }{3}$ + 2π3$\frac{2\pi }{3}$ = 0. Rewritten: π6$\frac{\pi }{6}$(4) − 2π3$\frac{2\pi }{3}$ = 2π3$\frac{2\pi }{3}$ − 2π3$\frac{2\pi }{3}$ = 0.The phase shift is the x where the inside is 0, which is where a cosine cycle starts. Both forms describe the same function, so both insides must be 0 there.
9. Check the range. At x = 4, y = 3 cos 0 − 1 = 3 − 1 = 2. Since cos runs from −1 to 1, y runs from 3(−1) − 1 = −4 to 3(1) − 1 = 2, so the range is [−4, 2].The range must be [B − |A|, B + |A|] = [−1 − 3, −1 + 3], centered on the midline. The cycle that starts at x = 4 must start at the top value 2, because cos 0 = 1.

Find the amplitude, period, phase shift and midline of y = 5 cos(−π8$\frac{\pi }{8}$x + 3π4$\frac{3\pi }{4}$) + 2. Say which way the graph is shifted. Then check the shift in both the original and the rewritten inside, and check the range. The figure is drawn from the answer and shows the midline and the five key points of one cycle.

1. The x-coefficient −π8$\frac{\pi }{8}$ is negative, so factor −1 out of the whole inside: −π8$\frac{\pi }{8}$x + 3π4$\frac{3\pi }{4}$ = −(π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$).The period and shift formulas need a positive speed ω. The −1 comes out of both terms, so the constant changes sign too: +3π4$\frac{3\pi }{4}$ becomes −3π4$\frac{3\pi }{4}$ inside the bracket. Expanding −(π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$) gives back −π8$\frac{\pi }{8}$x + 3π4$\frac{3\pi }{4}$, so the function has not changed.
2. Use the cosine even rule cos(−u) = cos u with u = π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$: y = 5 cos(π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$) + 2.Cosine is even. On the unit circle the angles u and −u are mirror images across the x-axis, so they share the same x-coordinate, and that x-coordinate is the cosine. The minus in front of the bracket drops and the 5 stays 5. Only a sine would change A, because sin(−u) = −sin u.
3. Match y = 5 cos(π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$) + 2 with y = A cos(ωx − φ) + B: A = 5, ω = π8$\frac{\pi }{8}$, φ = 3π4$\frac{3\pi }{4}$, B = 2.The pattern has a minus in front of φ, so the −3π4$\frac{3\pi }{4}$ in the bracket means φ = +3π4$\frac{3\pi }{4}$. Keeping that sign decides the direction of the shift.
4. Amplitude: |A| = |5| = 5.The amplitude is the vertical distance from the midline to a maximum, and a distance is never negative.
5. Period: 2πω$\frac{2\pi }{\omega }$ = 2π ÷ π8$\frac{\pi }{8}$ = 2π × 8π$\frac{8}{\pi }$ = 16.Dividing by the fraction π8$\frac{\pi }{8}$ means multiplying by its reciprocal 8π$\frac{8}{\pi }$. The π cancels, leaving 2 × 8 = 16. One full cycle covers 16 units of x.
6. Phase shift: φω$\frac{\phi }{\omega }$ = 3π4$\frac{3\pi }{4}$ ÷ π8$\frac{\pi }{8}$ = 3π4$\frac{3\pi }{4}$ × 8π$\frac{8}{\pi }$ = 244$\frac{24}{4}$ = 6. It is positive, so the graph is shifted 6 units to the right.Both positions hold fractions, so again multiply by the reciprocal of ω. The π cancels and 34$\frac{3}{4}$ × 8 = 6. In y = A cos(ωx − φ) + B a positive φω$\frac{\phi }{\omega }$ means a shift to the right. The factored form π8$\frac{\pi }{8}$x − 3π4$\frac{3\pi }{4}$ = π8$\frac{\pi }{8}$(x − 6) shows the same thing.
7. Midline: y = B = 2.The + 2 is added outside the cosine, so every point of the graph is lifted by 2 and the graph swings evenly about the line y = 2.
8. Plug the shift x = 6 into both insides. Original: −π8$\frac{\pi }{8}$(6) + 3π4$\frac{3\pi }{4}$ = −3π4$\frac{3\pi }{4}$ + 3π4$\frac{3\pi }{4}$ = 0. Rewritten: π8$\frac{\pi }{8}$(6) − 3π4$\frac{3\pi }{4}$ = 3π4$\frac{3\pi }{4}$ − 3π4$\frac{3\pi }{4}$ = 0. At that point y = 5 cos 0 + 2 = 7.A cosine cycle starts where its inside is 0. Both insides are 0 at the same x, which shows the rewrite kept the function the same. The value y = 7 = B + |A| shows the cycle starts at a maximum, as a cosine cycle should.
9. Check the range: cosine runs from −1 to 1, so y runs from 5(−1) + 2 = −3 to 5(1) + 2 = 7, giving −3 ≤ y ≤ 7.The graph reaches |A| above and |A| below the midline: 2 + 5 = 7 and 2 − 5 = −3. The top value 7 agrees with the maximum found at x = 6.

Find the amplitude, period, phase shift and midline of y = −2 cos(−π5$\frac{\pi }{5}$x − 3π5$\frac{3\pi }{5}$) − 4. Say which way the graph is shifted. Then check the shift in both the original and the rewritten inside, and check the range. The figure shows the finished result: the midline drawn as a horizontal line, with the lowest, highest and next lowest points of one cycle marked.

1. The x-coefficient inside is −π5$\frac{\pi }{5}$, which is negative, so factor −1 out of the whole inside: −π5$\frac{\pi }{5}$x − 3π5$\frac{3\pi }{5}$ = −(π5$\frac{\pi }{5}$x + 3π5$\frac{3\pi }{5}$).The period and shift formulas need a positive speed. The −1 has to come out of both terms, so the minus in front of 3π5$\frac{3\pi }{5}$ becomes a plus inside the bracket.
2. Use the cosine even rule cos(−u) = cos u with u = π5$\frac{\pi }{5}$x + 3π5$\frac{3\pi }{5}$: y = −2 cos(π5$\frac{\pi }{5}$x + 3π5$\frac{3\pi }{5}$) − 4.Cosine is even, so removing the −1 inside changes nothing outside. The −2 in front and the −4 at the end stay as they were. With sine, the odd rule would also flip the sign of A.
3. Match with y = A cos(ωx − φ) + B. The inside π5$\frac{\pi }{5}$x + 3π5$\frac{3\pi }{5}$ means ω = π5$\frac{\pi }{5}$ and −φ = 3π5$\frac{3\pi }{5}$. So A = −2, ω = π5$\frac{\pi }{5}$, φ = −3π5$\frac{3\pi }{5}$, B = −4.The form has a minus before φ, so a plus inside makes φ negative. Keeping every sign records three things: the graph is flipped (A < 0), it moves left (φ < 0), and the midline is below the x-axis (B < 0).
4. Amplitude = |A| = |−2| = 2.Amplitude is the distance from the midline to a peak, so it is never negative. The minus on A only turns the graph upside down.
5. Period = 2πω$\frac{2\pi }{\omega }$ = 2π ÷ π5$\frac{\pi }{5}$ = 2π · 5π$\frac{5}{\pi }$ = 10.After the rewrite ω = π5$\frac{\pi }{5}$ is positive, so the period is positive. One full cycle takes 10 units of x.
6. Phase shift = φω$\frac{\phi }{\omega }$ = −3π5$\frac{3\pi }{5}$ ÷ π5$\frac{\pi }{5}$ = −3π5$\frac{3\pi }{5}$ · 5π$\frac{5}{\pi }$ = −3, so the graph is shifted 3 units to the left.A negative phase shift means a move to the left. The minus in front of 3π5$\frac{3\pi }{5}$ in the original became a plus when −1 was factored out, and that plus is what sends the graph left.
7. Midline: y = B = −4.B is the vertical shift and keeps its sign, so the graph oscillates about the line y = −4, below the x-axis.
8. Put x = −3 into both insides. Original: −π5$\frac{\pi }{5}$(−3) − 3π5$\frac{3\pi }{5}$ = 3π5$\frac{3\pi }{5}$ − 3π5$\frac{3\pi }{5}$ = 0. Rewritten: π5$\frac{\pi }{5}$(−3) + 3π5$\frac{3\pi }{5}$ = −3π5$\frac{3\pi }{5}$ + 3π5$\frac{3\pi }{5}$ = 0. Then y = −2 cos 0 − 4 = −6. The range runs from −4 − 2 = −6 to −4 + 2 = −2.At the phase shift the rewritten inside is 0. The original inside is its negative, so it is 0 as well. Because A < 0, a cycle starts at its lowest point, and that lowest value has to be B − |A|.