An equation of the line tangent to the graph of y = 2x+3/3x-2 at the point (1,5) is

We want to find the equation of the tangent line to a given point in a given function.

The correct option is B: 13*x + y = 18

So we want to find the equation of the tangent line to the function.

at the point (1, 5).

To get the slope of that line, we need to evaluate the derivate of the function in the x-value of the point, the derivate of the function will be:

Now we evaluate this in the x-value of the point, which is x = 1.

Then the line will be something like:

y = -13*x + b

To find the value of b, we use the fact that the line must pass through the point (1, 5). This means that when x = 1, we have y = 5.

Replacing that in the line equation we get:

5 = -13*1 + b

5 + 13 = b

18 = b

Then the equation of the line is:

y = -13*x + 18

Or, rewriting it.

13*x + y = 18

So the correct option is B.

If you want to learn more, you can read:

link

13x + y = 18

Step-by-step explanation: 13x + y = 18 is equal to y = -13x + 18.

The derivative is y' = -13/3x-2^2.

With the derivative, you get -13 for slope and plug it in to get y - 5 = -13x + 13.

Move it around and you get 13x + y = 18.

Hope this helps!