\( q(x)=2+x-2 x^{2}-x^{3}, x \in \mathbb{R} \), is a cubic function. (a) Find each of the following: (i) \( q(-1) \) (ii) \( q^{\prime}(x) \) (iii) \( q^{\prime}(-1) \) b) \( y=q(x), x \in \mathbb{R} \), is a graph of \( q(x) \). (i) Show that \( (1,0) \) is a point on the graph of \( q(x) \). (ii) Find the slope of the tangent to \( y=q(x) \) at the point \( (1,0) \). (iii) Hence, find the equation of the tangent to \( y=q(x) \) at \( (1,0) \). Give your answer in the form \( a x+b y+c=0 \). [Hint: use \( y-y_{1}=m\left(x-x_{1}\right) \) ]

Let's analyze the cubic function \( q(x) = 2 + x - 2x^{2} - x^{3} \) and address each part of the problem step by step. --- ## **a) Calculations Involving \( q(x) \)** ### **i) Find \( q(-1) \)** To find \( q(-1) \), substitute \( x = -1 \) into the function: \[ \begin{align*} q(-1) &= 2 + (-1) - 2(-1)^{2} - (-1)^{3} \\ &= 2 - 1 - 2(1) - (-1) \\ &= 2 - 1 - 2 + 1 \\ &= (2 - 1) + (-2 + 1) \\ &= 1 - 1 \\ &= 0 \end{align*} \] **Answer:** \( q(-1) = 0 \) --- ### **ii) Find \( q^{\prime}(x) \)** To find the derivative \( q^{\prime}(x) \), differentiate \( q(x) \) with respect to \( x \): \[ \begin{align*} q(x) &= 2 + x - 2x^{2} - x^{3} \\ q^{\prime}(x) &= \frac{d}{dx}(2) + \frac{d}{dx}(x) - \frac{d}{dx}(2x^{2}) - \frac{d}{dx}(x^{3}) \\ &= 0 + 1 - 4x - 3x^{2} \\ &= -3x^{2} - 4x + 1 \end{align*} \] **Answer:** \( q^{\prime}(x) = -3x^{2} - 4x + 1 \) --- ### **iii) Find \( q^{\prime}(-1) \)** Using the derivative found in part (ii), substitute \( x = -1 \): \[ \begin{align*} q^{\prime}(-1) &= -3(-1)^{2} - 4(-1) + 1 \\ &= -3(1) + 4 + 1 \\ &= -3 + 4 + 1 \\ &= 2 \end{align*} \] **Answer:** \( q^{\prime}(-1) = 2 \) --- ## **b) Graph Analysis of \( y = q(x) \)** ### **i) Show that \( (1, 0) \) is a Point on the Graph of \( q(x) \)** To verify that \( (1, 0) \) lies on the graph of \( q(x) \), substitute \( x = 1 \) into \( q(x) \) and check if \( y = 0 \). \[ \begin{align*} q(1) &= 2 + (1) - 2(1)^{2} - (1)^{3} \\ &= 2 + 1 - 2(1) - 1 \\ &= 2 + 1 - 2 - 1 \\ &= (2 + 1) + (-2 -1) \\ &= 3 - 3 \\ &= 0 \end{align*} \] Since \( q(1) = 0 \), the point \( (1, 0) \) lies on the graph of \( q(x) \). **Answer:** \( (1, 0) \) is a point on the graph of \( q(x) \). --- ### **ii) Find the Slope of the Tangent to \( y = q(x) \) at \( (1, 0) \)** The slope of the tangent at a point \( x = a \) is given by the derivative \( q^{\prime}(a) \). From part **a(ii)**, we have: \[ q^{\prime}(x) = -3x^{2} - 4x + 1 \] Substitute \( x = 1 \): \[ \begin{align*} q^{\prime}(1) &= -3(1)^{2} - 4(1) + 1 \\ &= -3 - 4 + 1 \\ &= -6 \end{align*} \] **Answer:** The slope of the tangent at \( (1, 0) \) is \( m = -6 \). --- ### **iii) Find the Equation of the Tangent to \( y = q(x) \) at \( (1, 0) \)** We can use the point-slope form of a line: \[ y - y_{1} = m(x - x_{1}) \] Where: - \( m \) is the slope - \( (x_{1}, y_{1}) \) is the point on the line Given: - \( m = -6 \) - \( (x_{1}, y_{1}) = (1, 0) \) Substitute these values: \[ \begin{align*} y - 0 &= -6(x - 1) \\ y &= -6x + 6 \end{align*} \] To express this in the form \( a x + b y + c = 0 \), rearrange the equation: \[ \begin{align*} y &= -6x + 6 \\ 6x + y - 6 &= 0 \end{align*} \] **Answer:** An equation for the tangent is \( 6x + y - 6 = 0 \). ---