Key Features Of F(x) = (5/4)sin(x) + 1

When we look at the graph of a function, understanding its key features is like getting to know its personality. For the function $f(x) = \frac{5}{4} \sin(x) + 1$, these features tell us a lot about its behavior. Let's break down what makes this particular trigonometric function tick. We'll explore its highest and lowest points, its behavior over specific intervals, and how its shape is determined by the numbers in its equation. Understanding these elements will help us interpret the graph more effectively and appreciate the nuances of sinusoidal functions. We'll dive into concepts like amplitude, vertical shift, maximum and minimum values, and how the function behaves within a given range.

Amplitude and Vertical Shift: Shaping the Sinusoidal Wave

The amplitude and vertical shift are two fundamental components that dictate the overall shape and position of a sinusoidal function. For our function, $f(x) = \frac{5}{4} \sin(x) + 1$, the amplitude is represented by the coefficient multiplying the sine term, which is $\frac{5}{4}$. The amplitude tells us how far the wave oscillates from its center line. A larger amplitude means a taller wave, while a smaller amplitude means a flatter wave. In this case, $\frac{5}{4}$ indicates that the wave stretches $\frac{5}{4}$ units above and $\frac{5}{4}$ units below its horizontal midline. This value is always positive, as it represents a distance. The vertical shift is the constant term added to the function, which is $+1$ in our equation. This shift moves the entire graph up or down along the y-axis. Without this $+1$, the function $y = \frac{5}{4} \sin(x)$ would oscillate around the x-axis (y=0). However, the $+1$ shifts this entire oscillating pattern upward by 1 unit. This means the center line, or midline, of our function is not the x-axis, but the horizontal line $y=1$. These two parameters, amplitude and vertical shift, work together to define the range of the function and its position in the coordinate plane. They are crucial for sketching the graph accurately and for understanding its maximum and minimum values, which we will discuss next.

Determining the Maximum and Minimum Values

The maximum value of the function is the highest point the graph reaches, and the minimum value is the lowest point. For $f(x) = \frac{5}{4} \sin(x) + 1$, these values are directly influenced by the amplitude and the vertical shift. The sine function, $\sin(x)$, naturally oscillates between $-1$ and $1$. When we multiply $\sin(x)$ by the amplitude $\frac{5}{4}$, the range becomes $-\frac{5}{4}$ to $\frac{5}{4}$. Finally, adding the vertical shift of $+1$ shifts this entire range. So, the maximum value will occur when $\sin(x)$ is at its peak value of $1$. In this case, the maximum value is $f(x)_{max} = \left(\frac{5}{4}\right) \times (1) + 1 = \frac{5}{4} + 1 = \frac{9}{4}$. This means the highest point on the graph is $\frac{9}{4}$. Conversely, the minimum value will occur when $\sin(x)$ is at its lowest value of $-1$. The minimum value is $f(x)_{min} = \left(\frac{5}{4}\right) \times (-1) + 1 = -\frac{5}{4} + 1 = -\frac{1}{4}$. Therefore, the lowest point on the graph is $-\frac{1}{4}$. The range of the function $f(x)$ is thus from $-\frac{1}{4}$ to $\frac{9}{4}$, which can be expressed as the interval $\left[-\frac{1}{4}, \frac{9}{4}\right]$. These maximum and minimum values are critical landmarks on the graph, indicating the upper and lower bounds of the function's output.

The Maximum Value of the Function is $\boxed{\frac{9}{4}}$

The Minimum Value of the Function is $\boxed{-\frac{1}{4}}$

Behavior on the Interval $\left(0, \frac{\pi}{2}\right)$

Understanding the behavior of the function on a specific interval is key to grasping its local trends and characteristics. Let's focus on the interval $\left(0, \frac{\pi}{2}\right)$, which represents the first quadrant in the unit circle. On this interval, the basic sine function, $\sin(x)$, starts at $0$ (but not including $0$ as per the open interval), increases steadily, and reaches its peak value of $1$ at $x = \frac{\pi}{2}$. Since our function is $f(x) = \frac{5}{4} \sin(x) + 1$, it inherits this increasing behavior. As $x$ increases from just above $0$ towards $\frac{\pi}{2}$, $\sin(x)$ increases from just above $0$ towards $1$. Consequently, $\frac{5}{4} \sin(x)$ also increases from just above $0$ towards $\frac{5}{4}$. Adding the vertical shift of $1$ means that $f(x)$ increases from just above $1$ towards $\frac{9}{4}$. Therefore, on the interval $\left(0, \frac{\pi}{2}\right)$, the function $f(x)$ is strictly increasing. This means that for any two values $x_1$ and $x_2$ in this interval, if $x_1 < x_2$, then $f(x_1) < f(x_2)$. The slope of the tangent line to the curve will be positive throughout this interval, indicating that the function's value is rising as $x$ moves from left to right. This increasing trend is a fundamental characteristic of the sine function in the first quadrant and is preserved and amplified by the amplitude and vertical shift in our modified function. It's important to note that while the function is increasing, it does not reach its absolute maximum value within this open interval, as the maximum occurs at $x = \frac{\pi}{2}$ which is not included in $\left(0, \frac{\pi}{2}\right)$. The function approaches its maximum value as $x$ approaches $\frac{\pi}{2}$ from the left.

On the interval $\left(0, \frac{\pi}{2}\right)$, the function is increasing.

Understanding the Period and Phase Shift

While not explicitly asked for in the initial prompts, understanding the period and phase shift provides a more complete picture of a sinusoidal function. For $f(x) = \frac{5}{4} \sin(x) + 1$, the period is determined by the coefficient of $x$ inside the sine function. In this case, the coefficient is $1$. The standard period for $\sin(x)$ is $2\pi$. The formula for the period of a function in the form $A \sin(Bx + C) + D$ is $\frac{2\pi}{|B|}$. Since $B=1$ here, the period is $\frac{2\pi}{|1|} = 2\pi$. This means the function completes one full cycle of its wave pattern every $2\pi$ units along the x-axis. A phase shift occurs when there's a horizontal translation, indicated by a term added or subtracted inside the sine function (like $x-h$). In $f(x) = \frac{5}{4} \sin(x) + 1$, there is no term added or subtracted inside the sine function (i.e., the argument is just $x$), so there is no horizontal phase shift. The graph starts its cycle in a way that is standard for the sine function, beginning at its midline and increasing. If the function were, for example, $f(x) = \frac{5}{4} \sin(x - \frac{\pi}{2}) + 1$, it would have a phase shift of $\frac{\pi}{2}$ to the right, meaning its standard starting point would be shifted. The absence of a phase shift in our function simplifies its analysis and graphing, as its horizontal positioning is aligned with the basic sine wave.

Key Takeaways and Graphical Interpretation

In summary, the function $f(x) = \frac{5}{4} \sin(x) + 1$ is a modified sine wave with several distinct characteristics. Its amplitude of $\frac{5}{4}$ signifies a vertical stretch, making the wave taller than a standard sine wave. The vertical shift of $+1$ elevates the entire graph, centering it around the horizontal line $y=1$ instead of the x-axis. This results in a maximum value of $\frac{9}{4}$ and a minimum value of $-\frac{1}{4}$. The function exhibits an increasing behavior on the interval $\left(0, \frac{\pi}{2}\right)$, mirroring the first quadrant's behavior of the basic sine function. The period remains $2\pi$ and there is no phase shift, indicating a standard horizontal alignment for the sine wave's cycle. When visualizing this on a graph, you would see a wave that oscillates between $-\frac{1}{4}$ and $\frac{9}{4}$, with its midline at $y=1$. The wave would start at $(0, 1)$, increase to its peak at $(\frac{\pi}{2}, \frac{9}{4})$, decrease through $( \pi, 1)$ and $( \frac{3\pi}{2}, -\frac{1}{4})$, and return to the midline at $(2\pi, 1)$, completing one full cycle. These features collectively define the unique graphical representation and behavior of $f(x) = \frac{5}{4} \sin(x) + 1$, making it an excellent example for studying transformations of trigonometric functions.

For further exploration into trigonometric functions and their properties, you can refer to resources like Khan Academy's Trigonometry section.