Finding amplitude period and midline of a sine function

Key takeaway: This is a basic sinusoidal transformation problem. The coefficients in front of sine and inside the angle control the graph’s height, horizontal stretch, and vertical shift.

Key features of the sine model

For a function of the form

$r(x)=a\sin(bx)+d,$

the three main features are:

• Amplitude = $|a|$
• Period = $\frac{2\pi}{|b|}$
• Midline = $y=d$

Here,

$r(x)=2\sin\left(\frac{\pi}{4}x\right)+5.$

So $a=2$, $b=\frac{\pi}{4}$, and $d=5$.

Step-by-step calculation

The amplitude is

$|2|=2.$

This means the graph moves 2 units above and below its midline.

The period is

$\frac{2\pi}{\pi/4}=2\pi\cdot\frac{4}{\pi}=8.$

So one full cycle takes 8 units along the $x$-axis.

The midline is

$y=5.$

This tells you the graph is shifted upward by 5 units.

What these values mean on the graph

The amplitude controls vertical stretch, the period controls how quickly the wave repeats, and the midline sets the central horizontal line around which the sine curve oscillates.

A useful check is that the maximum value should be $5+2=7$, and the minimum value should be $5-2=3$. If those match the graph shape, your parameters are consistent.

For transformed sine functions, the fastest route is to identify the coefficient on sine, the coefficient of $x$ inside the function, and the constant shift. That gives the graph’s essential behavior without sketching every point.

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Pitfalls the pros know 👇 A common mistake is to think the period is $\frac{2\pi}{\pi/4}=\frac{8\pi}{\pi}$ and then leave the answer as something like $8\pi$. The $\pi$ cancels, and the period is a plain number, 8. Another error is using the outside number 5 as the amplitude. In sine graphs, the vertical shift is the midline, not the amplitude. The amplitude comes from the coefficient multiplying the sine function, even when the function is shifted upward or downward.

What if the problem changes? If the function were $r(x)=-2\sin\left(\frac{\pi}{4}x\right)+5$, the amplitude would still be 2 and the period would still be 8. The negative sign would only reflect the graph across the midline, not change its size or repetition rate. The midline would remain $y=5$. This variant shows why amplitude is always the absolute value of the coefficient in front of sine or cosine.

Tags: amplitude, period of a sine function, midline