Question

An equation of the line tangent to the graph of......(see attachment)

Answer 1

a) take the derivative
b) replace x by 1 to find your slope
c) use the point slope formula to find the equation of the line

let me know which of any you need help with

Answer 2

okay. I got: \[y = 5x-1(3) - 3x +1(5)\div (3x + )^2\]

Answer 3

for derivative that is

Answer 4

y= 4/16

Answer 5

y – y1 = m(x – x1)

Answer 6

If i'm right so far, can you help with point slope formula?

Answer 7

hold on let me check the derivative

Answer 8

did you get
\[y'=\frac{8}{(3x+1)^2}\]?

Answer 9

if you replace x by 1 in the original expression you get 1 for y , so your point is (1,1)
if you replace x by 1 in the derivative you get \(\frac{8}{16}=\frac{1}{2}\)

Answer 10

point slope formula you have written, it is
\[y-y_1=m(x-x_1)\] in this case
\[y-1=\frac{1}{2}(x-1)\]

Answer 11

i think maybe your derivative is wrong, but it is hard for me to read it

Answer 12

@satellite73 yes, mine is same as yours!

Answer 13

@needhlp
y' = [ 5 ( 3x +1 ) - 3 ( 5x -1) ] / ( 3x +1)^2