The slope of the tangent to the curve `x=t^2+3t-8,``y=2t^2-2t-5`at the point `(2, 1)`is(A) `(22)/7` (B) `6/7` (C) `7/6` (D) `(-6)/7`

Explanation

To find the slope of the tangent to the curve defined by the parametric equations \( x = t^2 + 3t - 8 \) and \( y = 2t^2 - 2t - 5 \) at the point \( (2, 1) \), we will follow these steps:

### Step 1: Find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

1. Differentiate \( y \) with respect to \( t \):
\[ y = 2t^2 - 2t - 5 \implies \frac{dy}{dt} = 4t - 2 \]
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