The angle between the tangents to the curve y | vedclass

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Question

(B) Given the curve $y = x^2 - 5x + 6$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 5$.At point $(2, 0)$,the slope of the tangent

Vedclass · Accepted Answer

$\pi /2$

Vedclass · Answer

(B) Given the curve $y = x^2 - 5x + 6$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 5$.At point $(2, 0)$,the slope of the tangent $m_1 = \left( \frac{dy}{dx} ight)_{(2, 0)} = 2(2) - 5 = -1$.At point $(3, 0)$,the slope of the tangent $m_2 = \left( \frac{dy}{dx} ight)_{(3, 0)} = 2(3) - 5 = 1$.Since $m_1 	imes m_2 = (-1) 	imes (1) = -1$,the tangents are perpendicular to each other.Therefore,the angle between the tangents is $\pi /2$.

The angle between the tangents to the curve $y = x^2 - 5x + 6$ at the points $(2, 0)$ and $(3, 0)$ is:

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