Problem 26
Question
Compare the graph of \(f(x)=4 / x^{2}\) with the graph of \(g\). $$g(x)=\frac{1}{8} f(x)=\frac{1}{2 x^{2}}$$
Step-by-Step Solution
Verified Answer
The graph of \(g(x)=\frac{1}{2x^{2}}\) is a vertically scaled down version of the graph of \(f(x)=4/x^{2}\). Both graphs have similar shapes and positions but the steepness of \(g(x)\) is less due to division by 8, which leads to a vertical scale down.
1Step 1: Understand the Function \(f(x)=4 / x^{2}\)
Before comparing, first understand the function \(f(x)=4 / x^{2}\). This is a basic function of a hyperbola with the graph opening upwards and downwards, and translated vertically, but scaled vertically by a factor of 4. A vertical scale of the factor changes the steepness of the graph, not the shape or position. Plotting it will give a more visual understanding.
2Step 2: Understand the Function \(g(x)=\frac{1}{2x^{2}}\)
This function is derived from \(f(x)\) by dividing it by 8 (where the 8 comes from \(4*2\), since the result is a function related to \(f(x)\) but divided by 2) which results in \(g(x)=\frac{1}{2x^{2}}\). This operation impacts the vertical scaling of \(f(x)\). The shape and position remain the same, but the graph is less steep, due to division by a positive constant larger than one.
3Step 3: Comparing the Graphs of \(f(x)\) and \(g(x)\)
If you plot both functions, you can visually see that both graphs have the same shape and position but differ in steepness. The graph of \(g(x)\) is less steep than the graph of \(f(x)\) because it's scaled down vertically. This is due to the \(f(x)\) function being divided by 8, making \(g(x)\)
Key Concepts
HyperbolaVertical ScalingFunction Comparison
Hyperbola
Graphs of hyperbolas take a distinct shape that appears in many mathematical contexts. A hyperbola is defined by its two branches, which bend away from each other. The general equation for a hyperbola involving a function like our exercise example is
In our example, the function \(f(x) = \frac{4}{x^2}\) represents a hyperbola where the opening is more pronounced due to the vertical scaling factor, which we will discuss next.
Hyperbolas are important in many fields because they model phenomena involving inverse square laws, such as gravity and light intensity.
- When the function is of the form \(f(x) = \frac{a}{x^2}\), it indicates a hyperbola that opens upward and downward.
In our example, the function \(f(x) = \frac{4}{x^2}\) represents a hyperbola where the opening is more pronounced due to the vertical scaling factor, which we will discuss next.
Hyperbolas are important in many fields because they model phenomena involving inverse square laws, such as gravity and light intensity.
Vertical Scaling
Vertical scaling affects how 'steep' a graph appears. This happens when a function is multiplied or divided by a constant. The graph's general shape and position remain the same, but its steepness—or how rapidly the y-values increase or decrease—changes.
Thus, vertical scaling can change how "spread out" or compressed a graph appears without altering the horizontal stretch or compress.
- In our context, \(f(x) = \frac{4}{x^2}\) is a vertically scaled version of the standard hyperbola \( \frac{1}{x^2} \) by a factor of 4.
- This means every point on the graph of \( \frac{1}{x^2} \) is stretched away from the x-axis by a factor of 4.
Thus, vertical scaling can change how "spread out" or compressed a graph appears without altering the horizontal stretch or compress.
Function Comparison
Comparing functions involves observing differences and similarities in their graphs. Key features to compare include steepness, position, and asymptotic behavior. In the exercise context, we compare \(f(x) = \frac{4}{x^2}\) and \(g(x) = \frac{1}{2x^2}\). Both functions have asymptotes aligning with the y-axis, indicative of their hyperbolic shape.
Here's how they compare:
Here's how they compare:
- Steepness: The graph of \(f(x)\) is steeper than that of \(g(x)\), due to the larger scaling factor of 4 in \(f(x)\) compared to a scaled-down value in \(g(x)\).
- Shape: Both graphs have the same hyperbolic shape, showing no change in curvature or direction.
- Position: Both are centered on the origin, maintaining identical positional attributes.
Other exercises in this chapter
Problem 26
Describe the right-hand and left-hand behavior of the graph of the polynomial function. $$f(s)=-\frac{2}{8}\left(s^{3}+5 s^{2}-7 s+1\right)$$
View solution Problem 26
Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$h(x)-x^{2}-4 x+2$$
View solution Problem 26
Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=9 x^{3}-15 x^{2}+11 x-5$$
View solution Problem 26
Perform the indicated operation and write the result in standard form. $$(1.6+3.2 i)+(-5.8+4.3 i)$$
View solution