y = -4 t^2 -t - 6
slope = dy/dt = -8 t - 1
at t = 8, slope = -65
so our tangent line is of form
y = -65 t + b
to find b, we need a point on the line
when t = 8,
y = -4(64) -8 -6
= -270
so use that point (8, -270) to find b
-270 = -65(8) + b
b = 250
so in the end
y = -65 t + 250
f(t) = − 4 t2 − t− 6
Find the equation of the line tangent to the graph of f(t) at t = 8.
Enter the equation of the tangent line here (in terms of the variable t):
1 answer