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 \( f \).

**(a)** To check if the point \((-2,15)\) is on the graph, substitute \( x = -2 \) into \( f(x) \): \[ f(-2)=4(-2)^2-(-2)-3 = 4(4)+2-3=16+2-3=15. \] Since \( f(-2)=15 \), the point \((-2,15)\) is on the graph. **(b)** Substitute \( x=2 \) into \( f(x) \): \[ f(2)=4(2)^2-2-3 =4(4)-2-3=16-2-3=11. \] Thus, when \( x=2 \), \( f(2)=11 \) and the point \((2,11)\) is on the graph. **(c)** To find \( x \) when \( f(x)=-3 \), solve: \[ 4x^2-x-3=-3. \] Add \( 3 \) to both sides: \[ 4x^2-x=0. \] Factor out \( x \): \[ x(4x-1)=0. \] Setting each factor to zero gives: \[ x=0 \quad \text{or} \quad 4x-1=0 \Rightarrow x=\frac{1}{4}. \] The points on the graph are \((0,-3)\) and \(\left(\frac{1}{4},-3\right)\). **(d)** The function \( f(x)=4x^2-x-3 \) is a polynomial. Therefore, its domain is all real numbers: \[ \text{Domain: } (-\infty,\infty). \] **(e)** The \( x \)-intercepts occur when \( f(x)=0 \): \[ 4x^2-x-3=0. \] Using the quadratic formula \[ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}, \quad \text{with } a=4, \, b=-1, \, c=-3, \] compute the discriminant: \[ \Delta=(-1)^2-4(4)(-3)=1+48=49. \] Thus, \[ x=\frac{-(-1) \pm \sqrt{49}}{2(4)}=\frac{1 \pm 7}{8}. \] The solutions are: \[ x=\frac{1+7}{8}=1 \quad \text{and} \quad x=\frac{1-7}{8}=-\frac{6}{8}=-\frac{3}{4}. \] So, the \( x \)-intercepts are \((1,0)\) and \(\left(-\frac{3}{4},0\right)\). **(f)** The \( y \)-intercept is found by evaluating \( f(0) \): \[ f(0)=4(0)^2-0-3=-3. \] Thus, the \( y \)-intercept is \((0,-3)\).