The function \( f \) is defined as follows. \( f(x)=\left\{\begin{array}{ll}2+2 x & \text { if } x<0 \\ x^{2} & \text { if } x \geq 0\end{array}\right. \) (a) Find the domain of the function. (b) Locate any intercepts. (c) Graph the function. (d) Based on the graph, find the range. (a) The domain of the function \( f \) is (Type your answer in interval notation.)

The domain of the function \( f \) is all real numbers, since the function has definitions for both \( x < 0 \) and \( x \geq 0 \). In interval notation, this is expressed as \( (-\infty, \infty) \). Now onto the intercepts! To find the y-intercept, set \( x = 0 \), which gives \( f(0) = 0^2 = 0 \). For the x-intercept, set \( f(x) = 0 \); in the case \( x \geq 0 \), we have \( x^2 = 0 \) leading to \( x = 0 \). In the piecewise section \( x < 0 \), \( 2 + 2x = 0 \) yields no x-intercepts since that would imply \( x < 0 \) but results in a negative value. Hence, the only intercept (both x and y) is at \( (0, 0) \). If you were to graph \( f \), you'd visually represent a line sloping upward to the left from \( (0, 2) \) as \( x \) approaches negative infinity, while the right side would show a parabolic curve starting at the origin and opening upwards. As for the range, based on this graph, you can see that \( f(x) \) takes on all values for \( y \) starting from 0 (at the vertex of the parabola) and then going upwards to infinity. Therefore, the range of the function is \( [0, \infty) \).