Given: \( f(x)=2 x^{\prime}-23 x^{2}+80 x-84 \) 43.1 Prove that \( (x-2) \) is a tactor of 43.2 Hence, or otherwisc, factorize \( f(x) \) fully 433 Determine the \( x \)-coordinates of the tuming points of \( f \) 434 Shetch the graph of / clearly labelling ALL. curning pounts and intercepls wath the axes. 435 Determine the coordinates of the -intercept of the tangent to of that has a slope of 40 and touches \( f \) at a point where the \( x \)-coordenate is integer NSC JUN 17

Let's solve each part of the problem step by step. ## Given Function: \[ f(x) = 2x^3 - 23x^2 + 80x - 84 \] ### **43.1 Prove that \((x - 2)\) is a factor of \(f(x)\)** **Using the Factor Theorem:** According to the Factor Theorem, \((x - c)\) is a factor of \(f(x)\) if and only if \(f(c) = 0\). **Calculate \(f(2)\):** \[ f(2) = 2(2)^3 - 23(2)^2 + 80(2) - 84 \] \[ f(2) = 2(8) - 23(4) + 160 - 84 \] \[ f(2) = 16 - 92 + 160 - 84 \] \[ f(2) = (16 - 92) + (160 - 84) = -76 + 76 = 0 \] **Conclusion:** Since \(f(2) = 0\), by the Factor Theorem, \((x - 2)\) is indeed a factor of \(f(x)\). --- ### **43.2 Factorize \(f(x)\) Fully** **Step 1: Factor Out \((x - 2)\) Using Synthetic Division** We already know that \(x = 2\) is a root. Let's perform synthetic division to factor out \((x - 2)\): \[ \begin{array}{r|rrrr} 2 & 2 & -23 & 80 & -84 \\ \hline & 2 & -19 & 42 & 0 \\ \end{array} \] **Result of Synthetic Division:** \[ f(x) = (x - 2)(2x^2 - 19x + 42) \] **Step 2: Factor the Quadratic \(2x^2 - 19x + 42\)** We need to factor \(2x^2 - 19x + 42\). **Find Two Numbers That Multiply to \(2 \times 42 = 84\) and Add to \(-19\):** - Numbers: \(-7\) and \(-12\) (since \(-7 \times -12 = 84\) and \(-7 + -12 = -19\)) **Rewrite and Factor by Grouping:** \[ 2x^2 - 7x - 12x + 42 = 2x(x - 3.5) - 12(x - 3.5) \] \[ = (2x - 12)(x - 3.5) \] \[ = 2(x - 6)(x - \frac{7}{2}) \] **Simplify:** \[ 2x^2 - 19x + 42 = (2x - 7)(x - 6) \] **Final Factorization of \(f(x)\):** \[ f(x) = (x - 2)(2x - 7)(x - 6) \] **Conclusion:** The fully factorized form of \(f(x)\) is: \[ f(x) = (x - 2)(2x - 7)(x - 6) \] --- ### **43.3 Determine the \(x\)-coordinates of the Turning Points of \(f(x)\)** **Finding the First Derivative:** \[ f'(x) = \frac{d}{dx}(2x^3 - 23x^2 + 80x - 84) = 6x^2 - 46x + 80 \] **Find Critical Points by Setting \(f'(x) = 0\):** \[ 6x^2 - 46x + 80 = 0 \] **Solve the Quadratic Equation:** \[ x = \frac{46 \pm \sqrt{(-46)^2 - 4 \times 6 \times 80}}{2 \times 6} = \frac{46 \pm \sqrt{2116 - 1920}}{12} = \frac{46 \pm \sqrt{196}}{12} \] \[ x = \frac{46 \pm 14}{12} \] \[ x = \frac{60}{12} = 5 \quad \text{or} \quad x = \frac{32}{12} = \frac{8}{3} \] **Conclusion:** The \(x\)-coordinates of the turning points are \(x = 5\) and \(x = \frac{8}{3}\). --- ### **43.4 Sketch the Graph of \(f(x)\) Clearly Labeling All Turning Points and Intercepts** *This section requires a graphical sketch, which cannot be provided in text form. However, here are the key features to plot:* - **Intercepts:** - **Y-intercept:** \(f(0) = -84\) \(\Rightarrow\) (0, -84) - **X-intercepts:** \(x = 2\), \(x = \frac{7}{2}\), and \(x = 6\) - **Turning Points:** - At \(x = 5\) and \(x = \frac{8}{3}\). Calculate corresponding \(f(x)\) values: - \(f(5) = 2(125) - 23(25) + 80(5) - 84 = 250 - 575 + 400 - 84 = -9\) \(\Rightarrow\) (5, -9) - \(f\left(\frac{8}{3}\right) = 2\left(\frac{512}{27}\right) - 23\left(\frac{64}{9}\right) + 80\left(\frac{8}{3}\right) - 84\) Simplify to find the exact or approximate value. - **End Behavior:** - As \(x \to \infty\), \(f(x) \to \infty\) - As \(x \to -\infty\), \(f(x) \to -\infty\) *Plot these points and behavior to sketch the graph.* --- ### **43.5 Determine the Coordinates of the Y-Intercept of the Tangent to \(f(x)\) with a Slope of 40 and Touching \(f(x)\) at a Point Where the \(x\)-Coordinate is Integer** **Step 1: Find the Derivative and Set It Equal to 40** \[ f'(x) = 6x^2 - 46x + 80 = 40 \] \[ 6x^2 - 46x + 40 = 0 \] **Solve for \(x\):** \[ 3x^2 - 23x + 20 = 0 \] \[ x = \frac{23 \pm \sqrt{529 - 240}}{6} = \frac{23 \pm \sqrt{289}}{6} = \frac{23 \pm 17}{6} \] \[ x = \frac{40}{6} = \frac{20}{3} \quad \text{or} \quad x = \frac{6}{6} = 1 \] **Only Integer \(x\)-Coordinate is \(x = 1\).** **Step 2: Find the Point of Tangency \((1, f(1))\):** \[ f(1) = 2(1)^3 - 23(1)^2 + 80(1) - 84 = 2 - 23 + 80 - 84 = -25 \] \[ \text{Point of Tangency: } (1, -25) \] **Step 3: Equation of the Tangent Line** Given: - Slope (\(m\)) = 40 - Point: (1, -25) Using point-slope form: \[ y - (-25) = 40(x - 1) \] \[ y + 25 = 40x - 40 \] \[ y = 40x - 65 \] **Y-Intercept of the Tangent Line:** Set \(x = 0\): \[ y = 40(0) - 65 = -65 \] \[ \text{Y-Intercept: } (0, -65) \] **Conclusion:** The tangent line with a slope of 40 touches \(f(x)\) at the integer \(x\)-coordinate \(x = 1\), and its y-intercept is at \((0, -65)\).