(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis. (b) The coordinates of the pin...
Question 1 Report
(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis.
(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).
Answer Details
(a) The curve \(y=(x^{2}-x+1)(x-2)\) cuts the x-axis where \(y=0\). Since \(x^{2}-x+1=0\) has discriminant \(1-4<0\) (no real roots), the only x-intercept is \(x-2=0\), i.e. the point \((2,0)\).
Expand: \(y=x^{3}-3x^{2}+3x-2\), so \(\dfrac{dy}{dx}=3x^{2}-6x+3\). At \(x=2\): \(3(4)-12+3=3\). This is the tangent gradient, so the normal gradient is \(-\tfrac13\).
Normal at \((2,0)\): \(y-0=-\tfrac13(x-2)\), i.e.
\[x+3y-2=0\]
(b) \(P(-1,2),\ Q(5,1),\ R(3,-4)\). Midpoint of \(\overline{PR}=\left(\tfrac{-1+3}{2},\tfrac{2-4}{2}\right)=(1,-1)\).
Line through \(Q(5,1)\) and \((1,-1)\): gradient \(=\dfrac{1-(-1)}{5-1}=\dfrac{2}{4}=\tfrac12\).
\[y-1=\tfrac12(x-5)\Rightarrow x-2y-3=0\]
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