It is given that curve x = t² + 3t – 8, y = 2t² – 2t – 5

Then,

The given points is (2, - 1)

At x = 2, we get

t² + 3t – 8 = 2

⇒ t² + 3t – 10 = 0

⇒ (t – 2)(t + 5) = 0

⇒ t = 2 and - 5

At y = - 1, we get

2t² - 2t – 5 = - 1

⇒ 2t² - 2t – 4 = 0

⇒ 2(t² - t – 2) = 0

⇒ (t – 2)(t + 1) = 0

⇒ t = 2 and - 1

Therefore, the common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, - 1) is