This is a very simple problem but I can't seem to get the correct answer for some reason.
Question: Find the slope of the function's graph at the given point.
h(t) = t^3 + 6t, (3, 45). Also. Find an equation for the line tangent to the graph there.
My answer is m=21 for the slope and for the equation of the tangent line, y=21x-54. Maybe I'm approaching the problem incorrectly?
I highly doubt it, but here it is anyways.
t^3-6t, Differentiates into f '(x) = 3t^2+6 . Factors into: 3(t^2+2)then I substituted 3 into t.
f(3)=3^3+6 = 33 . then i used the point slope formula to find the tangent line.
y-9 = 33(x-3) = y=33x-90
Ok, I got the slope correct, but my tanget line is still incorrect.