Answer the questions about the following function. \( f(x)=4 x^{2}-x-3 \) (a) Is the point \( (-2,15) \) on the graph of \( f \) ? (b) If \( x=2 \), what is \( f(x) \) ? What point is on the graph of \( f \) ? (c) If \( f(x)=-3 \), 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 \( (-2,15) \) is on the graph of \( f \), we need to evaluate \( f(-2) \). Plugging in \( -2 \): \[ f(-2) = 4(-2)^{2} - (-2) - 3 = 4(4) + 2 - 3 = 16 + 2 - 3 = 15 \] So, yes, the point \( (-2,15) \) is indeed on the graph of \( f \)! Now, for part (b), let's find \( f(2) \): \[ f(2) = 4(2)^{2} - (2) - 3 = 4(4) - 2 - 3 = 16 - 2 - 3 = 11 \] Therefore, the point \( (2,11) \) is on the graph of \( f \). Next, for part (c), if \( f(x) = -3 \): Setting the equation \( 4x^{2} - x - 3 = -3 \): \[ 4x^{2} - x - 3 + 3 = 0 \Rightarrow 4x^{2} - x = 0 \] Factoring gives us \( x(4x - 1) = 0 \), so \( x = 0 \) or \( x = \frac{1}{4} \). Thus, the points \( (0, -3) \) and \( \left(\frac{1}{4}, -3\right) \) are on the graph of \( f \). For part (d), the domain of \( f(x) \) is all real numbers since it's a polynomial function, which can take any x-value. In part (e), to find the \( x \)-intercepts, we set \( f(x) = 0 \): \[ 4x^{2} - x - 3 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 4 \), \( b = -1 \), and \( c = -3 \): \[ x = \frac{1 \pm \sqrt{(-1)^{2} - 4(4)(-3)}}{2(4)} = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8} \] So we get \( x = 1 \) and \( x = -\frac{3}{4} \). The \( x \)-intercepts are \( (1, 0) \) and \( \left(-\frac{3}{4}, 0\right) \). Finally, for part (f), the \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = 4(0)^{2} - 0 - 3 = -3 \] Thus, the \( y \)-intercept is \( (0, -3) \).