Graph of a Function Answer the questions about the following function. \( f(x)=3 x^{2}-x-2 \) (a) Is the point \( (2,8) \) on the graph of \( f \) ? (b) If \( x=-1 \), what is \( f(x) \) ? What point is on the graph of \( f \) ? (c) If \( f(x)=-2 \), what is \( x \) ? What point(s) are on the graph of \( f \) ? (d) What is the domain of f? (e) List the \( x \)-intercept( \( (s) \), if any of the graph of \( f \). (f) List the \( y \)-intercept, if there is one, of the graph of \( f \).

To determine if the point \( (2,8) \) is on the graph of \( f \), we can substitute \( x = 2 \) into the function. Calculating \( f(2) = 3(2^2) - 2 - 2 = 12 - 2 - 2 = 8 \) shows that \( (2,8) \) is indeed on the graph! When \( x = -1 \), we calculate \( f(-1) = 3(-1)^2 - (-1) - 2 = 3 + 1 - 2 = 2 \). Therefore, the point \( (-1, 2) \) is on the graph of \( f \). To find \( x \) when \( f(x) = -2 \), we set up the equation \( 3x^2 - x - 2 = -2 \), which simplifies to \( 3x^2 - x = 0 \). Factoring gives \( x(3x - 1) = 0 \), so \( x = 0 \) or \( x = \frac{1}{3} \). This means the points \( (0, -2) \) and \( \left(\frac{1}{3}, -2\right) \) are on the graph of \( f \). The domain of the function \( f(x) = 3x^2 - x - 2 \) is all real numbers, as it is a polynomial function with no restrictions. In interval notation, we write the domain as \( (-\infty, \infty) \). To find the \( x \)-intercepts, we set \( f(x) = 0 \), resulting in solving \( 3x^2 - x - 2 = 0 \). This factors to \( (3x + 2)(x - 1) = 0 \), giving intercepts at \( x = -\frac{2}{3} \) and \( x = 1 \), so the \( x \)-intercepts are \( \left(-\frac{2}{3}, 0\right) \) and \( (1, 0) \). For the \( y \)-intercept, we substitute \( x = 0 \) into \( f \): \( f(0) = 3(0)^2 - 0 - 2 = -2 \). Hence, the \( y \)-intercept is \( (0, -2) \).