Coordinate Geometry (Intersection/ Tangent)

CIE/2020/w/22/No 5

DO NOT USE A CALCULATOR IN THIS QUESTION.

(a) Find the equation of the tangent to the curve $y=x^{3}-6 x^{2}+3 x+10$ at the point where $x=1$.

(b) Find the coordinates of the point where this tangent meets the curve again.

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(a) $\begin{array}[t]{ll}&y=x^{3}-6 x^{2}+3 x+10 \\&\dfrac{d y}{d x}=3 x^{2}-12 x+3 \\&\text { when } x=1, y=(1)^{3}-6(1)^{2}+3(1)+10=8 \\&\dfrac{d y}{d x}=3(1)^{2}-12(1)+3=-6 \\&\text { tangent equation: } y-8=-6(x-1) \\&y=-6 x+14\end{array}$

(b) $\begin{array}[t]{rl}& x^{3}-6 x^{2}+3 x+10=-6 x+14 \\&x^{3}-6 x^{2}+9 x-4=0 \\&\text { Since } x=1 \text { is a solution }\\&x^{3}-6 x^{2}+9 x-4\\ &=(x-1)\left(x^{2}+b x+c\right) \\&=x^{3}+b x^{2}+c x-x^{2}-b x-c \\&=x^{3}+(b-1) x^{2}+(c-b) x-c \end{array}$

By equating coefficients, $c=4,b-1=-6,c-b=9.$ Thus $c=4,b=-5.$

$\begin{array}{rl}& \therefore x^{3}-6 x^{2}+9 x-4=(x-1)\left(x^{2}-5 x+4\right)=0 \\&(x-1)(x-1)(x-4)=0 \\&x=1 \text { or } x=4 \\&\text { When } x=4, y=(4)^{3}-6(4)^{2}+3(4)+10=-10 \\&\therefore(4,-10) \end{array}$

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