Find the slope of the tangents to the curve y | vedclass

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

Question

(C) Given the curve equation: $y = (x + 1)(x - 3)$.To find the points where the curve meets the $x$-axis,set $y = 0$:$0 = (x + 1)(x - 3)$This gives $x

Vedclass · Accepted Answer

$\pm 4$

Vedclass · Answer

(C) Given the curve equation: $y = (x + 1)(x - 3)$.To find the points where the curve meets the $x$-axis,set $y = 0$:$0 = (x + 1)(x - 3)$This gives $x = -1$ and $x = 3$.So,the points of intersection are $(-1, 0)$ and $(3, 0)$.Now,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$y = x^2 - 2x - 3$$\frac{dy}{dx} = 2x - 2$At point $(-1, 0)$:Slope $m_1 = 2(-1) - 2 = -2 - 2 = -4$.At point $(3, 0)$:Slope $m_2 = 2(3) - 2 = 6 - 2 = 4$.Thus,the slopes of the tangents at these points are $\pm 4$.

Find the slope of the tangents to the curve $y = (x + 1)(x - 3)$ at the points where it meets the $x$-axis.

Similar Questions

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

For the curve $y=4x^{3}-2x^{5}$,find all the points at which the tangent passes through the origin.

If the normal to the curve $y = f(x)$ is parallel to the $x-$axis,then the correct statement is:

If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other,then

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module