Consider the function \( f(x)=x^{3}+2 x^{2}-3 \). (a) Graph the function. (b) What are the \( x \) - and \( y \)-intercepts of the graph?

To graph the function \( f(x) = x^{3} + 2x^{2} - 3 \), we can start by finding the \( x \)-intercepts and the \( y \)-intercept. (a) To find the \( x \)-intercepts, we set \( f(x) = 0 \) and solve for \( x \): \[ x^{3} + 2x^{2} - 3 = 0 \] To find the \( y \)-intercept, we substitute \( x = 0 \) into the function: \[ f(0) = 0^{3} + 2(0)^{2} - 3 = -3 \] Therefore, the \( x \)-intercepts are the solutions to the equation \( x^{3} + 2x^{2} - 3 = 0 \), and the \( y \)-intercept is at the point (0, -3). (b) The \( x \)-intercepts are the values of \( x \) where the graph crosses the \( x \)-axis, and the \( y \)-intercept is the value of \( y \) where the graph crosses the \( y \)-axis. Let's calculate the \( x \)-intercepts by solving the equation \( x^{3} + 2x^{2} - 3 = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}+2x^{2}-3=0\) - step1: Factor the expression: \(\left(x-1\right)\left(x^{2}+3x+3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-1=0\\&x^{2}+3x+3=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=1\\&x \notin \mathbb{R}\end{align}\) - step4: Find the union: \(x=1\) The \( x \)-intercepts of the graph are at \( x = 1 \). Now, let's calculate the \( y \)-intercept by substituting \( x = 0 \) into the function \( f(x) = x^{3} + 2x^{2} - 3 \). Evaluate the expression by following steps: - step0: Evaluate: \(x^{3}+2x^{2}-3\) - step1: Substitute: \(0^{3}+2\times 0^{2}-3\) - step2: Calculate: \(0+2\times 0^{2}-3\) - step3: Calculate: \(0+2\times 0-3\) - step4: Multiply: \(0+0-3\) - step5: Remove 0: \(-3\) The \( y \)-intercept of the graph is at \( y = -3 \). Therefore, the \( x \)-intercepts are at \( x = 1 \) and the \( y \)-intercept is at \( y = -3 \).