Question

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there.
h(t)= t^3+3t at (1,4)

Answer 1

calculus class right?

Answer 2

yup

Answer 3

take the derivative of \(h(t)=t^3+3t\) by the power rule
what do you get?

Answer 4

\[3t ^{2}+3\]

Answer 5

ok good, so \(h'(t)=3t^2+3\)
to find your slope at \((1,4)\) compute \(h'(1)\)

Answer 6

then use the point slope formula to find the equation of the line through \((1,4)\) with the slope you find via \(h'(1)\)

Answer 7

so the slope is 6?

Answer 8

yes, the derivative is a formula for finding the slope of the tangent line

Answer 9

then \(y-y_1=m(x-x_1) \) with \(m=6,x_1=1,y_1=4\) gives you the equation of the line

Answer 10

so y=6x+10

Answer 11

that is not what i get

Answer 12

\[y-4=6(x-1)\]

Answer 13

you might have put \(6(x+1)\) by mistake

Answer 14

oh I did, now I got it

Answer 15

good luck!

Answer 16

Thanks so much!