Consider the given function \( d(x)=3 x^{2}-12 x+11 \). (a) Write the function in vertex form. (b) Identify the vertex. (c) Determine the \( x \)-intercept(s). (d) Determine the \( y \)-intercept(s). (e) Sketch the function. (f) Determine the axis of symmetry. (g) Determine the minimum or maximum value of the function. (h) Write the domain and range in interval notation. Write your answers in exact form.

**(a) Vertex Form** Start with the given function \[ d(x)=3x^2-12x+11. \] Factor out the coefficient of \(x^2\) from the quadratic and linear terms: \[ d(x)=3(x^2-4x)+11. \] Complete the square for \(x^2-4x\). Recall that \[ x^2-4x=(x-2)^2-4. \] Substitute this back into the function: \[ \begin{aligned} d(x)&=3\left[(x-2)^2-4\right]+11\\[1mm] &=3(x-2)^2-12+11\\[1mm] &=3(x-2)^2-1. \end{aligned} \] So, the vertex form is \[ d(x)=3(x-2)^2-1. \] --- **(b) Vertex** The vertex form \(d(x)=3(x-2)^2-1\) shows that the vertex is at \[ (2,-1). \] --- **(c) \(x\)-Intercept(s)** Set \(d(x)=0\) to find the \(x\)-intercepts: \[ 3x^2-12x+11=0. \] Use the quadratic formula, \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, \] where \(a=3\), \(b=-12\), and \(c=11\): \[ x=\frac{12\pm\sqrt{(-12)^2-4\cdot3\cdot11}}{2\cdot3}=\frac{12\pm\sqrt{144-132}}{6}=\frac{12\pm\sqrt{12}}{6}. \] Simplify \(\sqrt{12}\) as \[ \sqrt{12}=2\sqrt{3}, \] so: \[ x=\frac{12\pm2\sqrt{3}}{6}=\frac{6\pm\sqrt{3}}{3}. \] The \(x\)-intercepts are \[ \left(\frac{6+\sqrt{3}}{3},\,0\right) \quad \text{and} \quad \left(\frac{6-\sqrt{3}}{3},\,0\right). \] --- **(d) \(y\)-Intercept(s)** To find the \(y\)-intercept, set \(x=0\): \[ d(0)=3(0)^2-12(0)+11=11. \] Thus, the \(y\)-intercept is \[ (0,\,11). \] --- **(e) Sketch of the Function** - The vertex is at \((2,-1)\). - The parabola opens upward (since the coefficient \(3>0\)). - The \(x\)-intercepts are \(\left(\frac{6\pm\sqrt{3}}{3},\,0\right)\). - The \(y\)-intercept is \((0,11)\). - The axis of symmetry is the vertical line \(x=2\). Plot these key points and the axis of symmetry, then draw a smooth curve passing through them. --- **(f) Axis of Symmetry** The axis of symmetry is the vertical line through the vertex: \[ x=2. \] --- **(g) Minimum or Maximum Value** Since the parabola opens upward, the vertex represents the minimum point. The minimum value of the function is \[ -1, \] which occurs at \(x=2\). --- **(h) Domain and Range in Interval Notation** - The domain of any quadratic function is all real numbers: \[ (-\infty,\infty). \] - Since the minimum value is \(-1\) and the parabola opens upward, the range is: \[ [-1,\infty). \]