Question

The coordinates of the points(s) at which the tangents to the curve $y=x^3-3x^2-7x+6$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by

• A. $(-1,9)$
• B. $(3,-15)$
• C. $(1,-3)$
• D. none

Answer 1

The correct option is B $(3,-15)$

Let the point of tangency be

$(x_1+y_1)$

Then the equation of tangent is

$\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$

when

$y=0,x=x_1+\frac{-y}{3x_1-6x_1-7}$

when

$x=0,y=y_1-x_1(3x_1-6x_1-7)$

$-2(x_1+\frac{-y}{3x_1-6x_1-7})=y_1-x_1(3x_1-6x_1-7)$

$-2\left(\frac{x_1(3x_1-6x_1-7)-y_1}{3x_1-6x_1-7}\right)=y_1-x_1(3x_1-6x_1-7)$

$2=3x_1-6x_1-7$

$3x_1-6x_1-9=0$

$x_1=-1or3$

when

$x_1=-1,y_1=(-1{)}^3-3(-1{)}^2-7(-1)+6=9$

and the y-intercept is

$9-(-1)\left[3\right(-1{)}^2-6(-1)-7]=11>0$

when

$x_1=3,y_1=(3{)}^3-3(3{)}^2-7\left(3\right)+6=-15$

and the y-intercept is

$-15-\left(3\right)\left[3\right(3{)}^2-6\left(3\right)-7]=-21<0$

So,the point of tangency is $(-1,9)$, this is the correct answer.