What is the angle of inclination of the tange | vedclass

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

Question

(C) Given the equation of the curve is $y = (x - 1)(x - 2) = x^2 - 3x + 2$.To find the slope of the tangent,we differentiate with respect to $x$:$\fra

Vedclass · Accepted Answer

$3\pi /4$

Vedclass · Answer

(C) Given the equation of the curve is $y = (x - 1)(x - 2) = x^2 - 3x + 2$.To find the slope of the tangent,we differentiate with respect to $x$:$\frac{dy}{dx} = 2x - 3$.Now,evaluate the slope at the point $(1, 0)$:$m = \left( \frac{dy}{dx} ight)_{(1, 0)} = 2(1) - 3 = -1$.The slope of the tangent is equal to $	an(\psi)$,where $\psi$ is the angle of inclination with the $x$-axis.$	an(\psi) = -1$.Since $	an(\psi) = -1$,the angle $\psi = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

What is the angle of inclination of the tangent to the curve $y = (x - 1)(x - 2)$ at the point $(1, 0)$ with the $x$-axis?

Similar Questions

If the tangent to the curve $y = \frac{x}{x^2-3}$,$x \in R, (x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0,0)$ on it,is parallel to the line $2x + 6y - 11 = 0$,then

Find the slope of the tangent to the curve $y = x^{3} - x + 1$ at the point where the $x$-coordinate is $2$.

Find a point on the curve $y=(x-2)^{2}$ at which the tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$.

If the slope of the tangent to the curve $y = x^3$ at a point is equal to the $y$-coordinate of that point,then the point is:

The angle between the curves $x^2=8y$ and $xy=8$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module