What is the angle between the tangents to the | vedclass

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(A) Given the curve $y = x^2 - 5x + 6$. First,find the derivative: $\frac{dy}{dx} = 2x - 5$. At point $(2, 0)$,the slope of the tangent $m_1 = \left.

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) Given the curve $y = x^2 - 5x + 6$. First,find the derivative: $\frac{dy}{dx} = 2x - 5$. 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 $m_2 = \left. \frac{dy}{dx} ight|_{(3,0)} = 2(3) - 5 = 1$. Let $	heta$ be the angle between the tangents. The formula is $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values: $	an 	heta = \left| \frac{-1 - 1}{1 + (-1)(1)} ight| = \left| \frac{-2}{0} ight| = \infty$. Since $	an 	heta = \infty$,the angle $	heta = \frac{\pi}{2}$.

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

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