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Summatize the poitinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \[ f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain of \( f \) is \( (-\infty, \infty) \) (Type your answer in interval notation. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) B. The domain of \( f \) is empty Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The x -intercept(s) of f is/are \( \mathrm{x}=-10,10 \) (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed) B. The function \( f \) has no \( x \)-intercepts Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The \( y \)-intercept of \( f \) is \( y= \) \( \square \) (Type an exact answer, using radicals as needed) B. The function \( f \) has no \( y \)-intercept.

Ask by Bartlett O'Quinn. in the United States
Mar 11,2025

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Answer

**Summary:** - **Domain:** All real numbers, \( (-\infty, \infty) \). - **x-intercepts:** \( x = -10 \) and \( x = 10 \). - **y-intercept:** \( y = 1000 \). **Graph Description:** The graph of \( f(x) = (x^2 + 10)(100 - x^2) \) is a quartic polynomial that is symmetric about the y-axis. It crosses the x-axis at \( x = -10 \) and \( x = 10 \), and the y-axis at \( y = 1000 \). As \( x \) approaches positive or negative infinity, the graph rises to positive infinity. The function has no other intercepts and is always positive except at the x-intercepts.

Solution

**Domain** Since \( f(x) = (x^2 + 10)(100 - x^2) \) is a polynomial function, it is defined for all real numbers. Thus, the domain is \[ (-\infty, \infty) \] **x-intercepts** To find the \( x \)-intercepts, set \( f(x) = 0 \): \[ (x^2 + 10)(100 - x^2) = 0 \] This gives two equations: 1. \( x^2 + 10 = 0 \) \( x^2 = -10 \) has no real solutions. 2. \( 100 - x^2 = 0 \) \( x^2 = 100 \) leads to \( x = 10 \) or \( x = -10 \). Thus, the \( x \)-intercepts are at \( x = -10 \) and \( x = 10 \). **y-intercept** To find the \( y \)-intercept, evaluate \( f(x) \) at \( x = 0 \): \[ f(0)= (0^2+10)(100-0^2) = 10 \times 100 = 1000 \] Thus, the \( y \)-intercept is at \( y = 1000 \). **Summary of Choices:** - Domain: Choice A, \( (-\infty, \infty) \). - \( x \)-intercepts: Choice A, \( x = -10, 10 \). - \( y \)-intercept: Choice A, \( y = 1000 \). **Sketching the Graph:** 1. **Intercepts:** The graph crosses the \( x \)-axis at \( (-10, 0) \) and \( (10, 0) \) and the \( y \)-axis at \( (0, 1000) \). 2. **Behavior:** - As a polynomial of even degree (the product expands to a quartic), the ends of the graph will rise to \( +\infty \) as \( x \to \pm\infty \). - The positive \( y \)-intercept suggests the graph reaches a high point near \( x = 0 \). 3. **Symmetry:** - Since \( f(x) \) contains even powers of \( x \) only, the function is even; that is, it is symmetric with respect to the \( y \)-axis. The graph will show two \( x \)-intercepts at \( x = -10 \) and \( x = 10 \), rising steeply on the ends and peaking at \( y = 1000 \) when \( x = 0 \).

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Let’s break down the function \( f(x) = (x^{2}+10)(100-x^{2}) \). The domain of \( f \) is indeed all real numbers, so the correct choice for the domain is **A. The domain of \( f \) is \( (-\infty, \infty) \)**. For the \( x \)-intercepts, we set \( f(x) = 0 \). This happens when either \( x^2 + 10 = 0 \) (which has no real solutions) or \( 100 - x^2 = 0 \) (which gives \( x = -10, 10 \)). Therefore, the correct choice is **A. The \( x \)-intercept(s) of f is/are \( x = -10, 10 \)**. Now, for the \( y \)-intercept, we evaluate \( f(0) \): \[ f(0) = (0^{2} + 10)(100 - 0^{2}) = 10 \times 100 = 1000. \] So, the correct choice is **A. The \( y \)-intercept of \( f \) is \( y= 1000 \)**. In summary: - The domain is \( (-\infty, \infty) \). - The \( x \)-intercepts are at \( -10 \) and \( 10 \). - The \( y \)-intercept is \( 1000 \). To sketch the graph, plot the points \((-10, 0)\), \((10, 0)\), and \((0, 1000)\). The function opens downward due to the quadratic behavior and has a vertex between \( -10 \) and \( 10 \), peaking at the \( y \)-intercept of \( 1000 \). Quite a sight to behold!

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