What is the slope of the tangent to the curve | vedclass

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(C) Given the parametric equations of the curve: $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,find the derivatives with respect to $t$: $\frac{dx

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$6/7$

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(C) Given the parametric equations of the curve: $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ $\frac{dy}{dt} = 4t - 2$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$. To find the value of $t$ at the point $(2, -1)$,substitute the coordinates into the equations: $2 = t^2 + 3t - 8 \Rightarrow t^2 + 3t - 10 = 0 \Rightarrow (t+5)(t-2) = 0 \Rightarrow t = 2, -5$. $-1 = 2t^2 - 2t - 5 \Rightarrow 2t^2 - 2t - 4 = 0 \Rightarrow t^2 - t - 2 = 0 \Rightarrow (t-2)(t+1) = 0 \Rightarrow t = 2, -1$. The common value is $t = 2$. Now,substitute $t = 2$ into the slope formula: $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

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

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