If the curves y = x^3 - 3x^2 - 8x - 4 and y = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the points of intersection,we equate the two curves: $x^3 - 3x^2 - 8x - 4 = 3x^2 + 7x + 4$. This simplifies to $x^3 - 6x^2 - 15x - 8 = 0$.

Vedclass · Accepted Answer

$x - y + 1 = 0$

Vedclass · Answer

(A) To find the points of intersection,we equate the two curves: $x^3 - 3x^2 - 8x - 4 = 3x^2 + 7x + 4$. This simplifies to $x^3 - 6x^2 - 15x - 8 = 0$. Factoring the cubic equation,we get $(x + 1)^2(x - 8) = 0$. The curves touch where the derivative values are equal and the points of intersection are repeated. At $x = -1$,the curves touch. Substituting $x = -1$ into $y = 3x^2 + 7x + 4$,we get $y = 3(-1)^2 + 7(-1) + 4 = 3 - 7 + 4 = 0$. So,the point of contact $P$ is $(-1, 0)$. Now,find the slope of the tangent at $P(-1, 0)$ by differentiating $y = 3x^2 + 7x + 4$: $\frac{dy}{dx} = 6x + 7$. At $x = -1$,the slope $m = 6(-1) + 7 = 1$. The equation of the tangent line at $(-1, 0)$ with slope $m = 1$ is $y - 0 = 1(x - (-1))$,which simplifies to $y = x + 1$,or $x - y + 1 = 0$.

If the curves $y = x^3 - 3x^2 - 8x - 4$ and $y = 3x^2 + 7x + 4$ touch each other at a point $P$,then the equation of the common tangent at $P$ is:

Similar Questions

What is the length of the tangent to the curve $x = a(\cos t + \log \tan(t/2))$,$y = a \sin t$ at any point?

If the line $ax + by + c = 0, ab \neq 0,$ is a tangent to the curve $xy = 1 - 2x,$ then

For the curve $b{y^2} = {(x + a)^3}$,the square of the subtangent is proportional to:

Find the slope of the tangent to the curve $y=\frac{x-1}{x-2}, x \neq 2$ at $x=10$.

The slope of the normal to the curve $y=\frac{x}{x^2+1}$ at $x=-4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module