Find the equations of the tangent line at t = 0
x = 2t^3 - 4t + 7, y = t + ln(t + 1)
x = 2t^3 - 4t + 7, y = t + ln(t + 1)
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x(0) = 7 and y(0) = 0, so the point is (7,0).
dx/dt = 6t² - 4 ==> dx/dt = -4 when t = 0
dy/dt = 1 + 1/(t + 1) ==> dy/dt = 2 when t = 0.
The slope is m = 2/(-4) = -1/2.
y = (-1/2)(x - 7).
dx/dt = 6t² - 4 ==> dx/dt = -4 when t = 0
dy/dt = 1 + 1/(t + 1) ==> dy/dt = 2 when t = 0.
The slope is m = 2/(-4) = -1/2.
y = (-1/2)(x - 7).