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Find the slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at t = 2.

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Slope of the tangent =\(\frac{dy}{dx}\)

\(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)\(\frac{4t-2}{2t+3}\)

\(\frac{dy}{dx}\) at t = 2

\((\frac{4t-2}{2t+3})_{at\,t = 2}\)

\(\frac{4(2)-2}{2(2)+3}\) 

\(\frac{6}{7}\)

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