At what point on the curve

Question

x = 3t2 + 7,y = t3 − 8
does the tangent line have slope 1/2

Answer 1

dy/dt = 3t^2
dx/dt = 6t

(dy/dt) / (dx/dt) = dy/dx = 3t^2/6t = t/2

so t/2 = 1/2 ===> t = 1
then:
x = 3(1)^2 + 7 = 10
y = 1^3 - 8 = -7

your point is (10,-7)