Thread: Find the equation of the tangent line

1. Find the equation of the tangent line

Find the equation of the tangent line to the graph of the function at z = 3.

f(z)= (4z^2)/(5z^2+2z)

I found the slope to be (8/289)

Do I just plug in 3 to the original equation, and use whatever y I get to be...

y-(whatever I get)=(8/289)(x-3)?

Thanks,

-Tyler

2. equation of line

$f(z)=(4z^2)/(5z^2+2z)$
$f'(z)=8/(25*(z^2)+(20*z)+4)$ (through use of partial fractions)
$f'(3)=8/(25*(3^2)+(20*3)+4)$
$f'(3)=8/289$
then find the y coordinate to accompany x

$y=(4z^2)/(5z^2+2z)$
$y=(4*(3^2)/(5*(3^2)+(2*3))$
$y=0.71$(approximation)
giving the coordinate $(3,0.71)$
then use $y-b=m(x-a)$
$y-0.71=8/289(x-3)$
giving $y=(8/289)x-0.97966$
rounding to $y=(8/289)x-1$