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Consider the following function. \[ f(x)=\left\{\begin{array}{rrr}5 x-2 & \text { if } & x=1 \\ x^{2}+1 & \text { if } & 15 x<11 \\ 7 & \text { if } & x \geq 11\end{array}\right. \] Which "piece(s)" should be used to find the \( y \)-intercept? Select all that apply. \( \square \) A. 7 \( \square \) B. \( x^{2}+1 \) \( \square \) C. \( 5 x-2 \)

Ask by Carrillo Bowers. in the United States
Mar 16,2025

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Tutor-Verified Answer

Answer

To find the \( y \)-intercept, use the piece \( x^2 + 1 \).

Solution

For the \( y \)-intercept we need the value of \( f(0) \). 1. First, note that the \( y \)-intercept occurs when \( x=0 \). 2. Check the conditions of the piecewise function: - The first piece applies only when \( x=1 \), so it does not apply for \( x=0 \). - The second piece applies when \( 15x < 11 \). Solving for \( x \): \[ 15x < 11 \implies x < \frac{11}{15} \] Since \( 0 < \frac{11}{15} \), the condition is satisfied. - The third piece applies when \( x \geq 11 \), which is not the case when \( x=0 \). 3. Therefore, for \( x=0 \), only the second piece \( x^2+1 \) applies. Thus, the correct piece to use for finding the \( y \)-intercept is: \( \square \) B. \( x^2+1 \)

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To find the \( y \)-intercept of a function, you need to evaluate the function at \( x = 0 \). In this case, we substitute \( x = 0 \) into the relevant pieces of the function: 1. For \( x = 0 \): - The condition \( 15x < 11 \) holds true (since \( 0 < 11 \)), so we use the piece \( x^2 + 1 \), which gives \( 0^2 + 1 = 1 \). 2. The piece \( 5x - 2 \) is only applicable when \( x = 1 \), so it does not contribute to the \( y \)-intercept. 3. The piece \( 7 \) is referenced for \( x \geq 11 \), which does not apply at \( x = 0 \). So, only the piece \( B \) (\( x^2 + 1 \)) is used to find the \( y \)-intercept, giving a value of \( 1 \). Now, you can mark the following: \( \square \) A. 7 \( \checked \) B. \( x^{2}+1 \) \( \square \) C. \( 5 x-2 \)

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