Question: Points: 0.6 of 1 (a) Find the domain of the function. (b) Locate any intercepts. (c) Choose the…

Points: 0.6 of 1

(a) Find the domain of the function. (b) Locate any intercepts. (c) Choose the correct graph of the function. (d) Based on the graph, find the range.

f(x) = \[ \begin{cases} 4 + x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} \]

B. There are no x-intercepts.

Locate any y-intercepts. Select the correct choice below and fill in any answer boxes within your choice.

A. The y-intercept(s) is/are 0. (Type an integer or a simplified fraction. Use a comma to separate answers as needed.)

B. There are no y-intercepts.

(c) Choose the correct graph of the function.

A. [Graph A] B. [Graph B] C. [Graph C]

Solution

(a) Find the domain of the function. The function is defined as: \[ f(x) = \begin{cases} 4 + x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} \] The domain of \( f(x) \) is all real numbers because there are no restrictions on \( x \) in either case. (b) Locate any intercepts. For the y-intercept, set \( x = 0 \): \[ f(0) = 0^2 = 0 \] Thus, the y-intercept is 0. For x-intercepts, solve \( f(x) = 0 \): For \( 4 + x = 0 \), solve: \[ x = -4 \] For \( x^2 = 0 \), solve: \[ x = 0 \] The x-intercepts are \( x = -4 \) and \( x = 0 \). (c) Choose the correct graph of the function. The correct graph should have a linear segment from negative x-values converging at \( x = -4 \) and a parabolic shape starting at the origin and opening upwards. From the options given, graph (A) matches this description. (d) Based on the graph, find the range. The range of \( f(x) \) starts from the lowest value of the graph to the highest: - For \( x < 0 \), the graph extends upwards without bound. - For \( x \geq 0 \), it is a parabola opening upwards starting at 0. Thus, the range of \( f(x) \) is all non-negative real numbers, \( [0, \infty) \).