The slope of the tangent at (1, 2) to the cur | vedclass

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(A) Given the parametric equations of the curve are $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. To find the value of $t$ at the point $(1, 2)$, we set

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$\frac{8}{5}$

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(A) Given the parametric equations of the curve are $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. To find the value of $t$ at the point $(1, 2)$, we set $x = 1$ and $y = 2$. For $x = 1$: $t^2 - 7t + 7 = 1 \Rightarrow t^2 - 7t + 6 = 0 \Rightarrow (t - 6)(t - 1) = 0$, so $t = 1$ or $t = 6$. For $y = 2$: $t^2 - 4t - 10 = 2 \Rightarrow t^2 - 4t - 12 = 0 \Rightarrow (t - 6)(t + 2) = 0$, so $t = 6$ or $t = -2$. The common value is $t = 6$. Now, we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$. At $t = 6$: $\frac{dx}{dt} = 2(6) - 7 = 12 - 7 = 5$. $\frac{dy}{dt} = 2(6) - 4 = 12 - 4 = 8$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8}{5}$.

The slope of the tangent at $(1, 2)$ to the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$ is:

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