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

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(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

$\pm 4$

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(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.

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