The slope of the tangent to the curve x = t^2 | vedclass

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(D) Given $x = t^2 + 3t - 8 = 2$ $t^2 + 3t - 10 = 0$ $(t + 5)(t - 2) = 0$ $t = -5$ or $t = 2$ Also $y = 2t^2 - 2t - 5 = -1$ $2t^2 - 2t - 4 = 0$ $t^2 -

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$6/7$

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(D) Given $x = t^2 + 3t - 8 = 2$ $t^2 + 3t - 10 = 0$ $(t + 5)(t - 2) = 0$ $t = -5$ or $t = 2$ Also $y = 2t^2 - 2t - 5 = -1$ $2t^2 - 2t - 4 = 0$ $t^2 - t - 2 = 0$ $(t - 2)(t + 1) = 0$ $t = 2$ or $t = -1$ Since the point $(2, -1)$ must satisfy both equations,we take the common value $t = 2$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $\frac{dx}{dt} = 2t + 3$ $\frac{dy}{dt} = 4t - 2$ $\frac{dy}{dx} = \frac{4t - 2}{2t + 3}$ At $t = 2$: $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

The slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:

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