# Equation of tangent line to curve

1. Dec 13, 2011

1. The problem statement, all variables and given/known data

Find all x coordinates of points (x, y) on the curve y = (x − 2)5 /(x − 4)3 where the tangent line is horizontal

2. Relevant equations

Quotient Rule - Differentiation

3. The attempt at a solution

So I could guess, that if tangent line is horizontal, the equation of tangent line to curve is y = c, where gradient is 0.

So, first derivative of the curve is suppose equal to zero.

dy/dx;
0 = ((5)(x - 2)4(1)(x - 4)3 - (x - 2)5(3)(x - 4)2(1)) / ((x - 4)3)2
0 = (5(x - 2)4(x - 4)3 - 3(x - 2)5(x - 4)2) / (x - 4)6

Okay now I'm stuck.

2. Dec 13, 2011

### Staff: Mentor

That's a good start.

What you need to do is to write the numerator in a factored form. The best way to do that is to find the common factors in 5(x - 2)4(x - 4)3 - 3(x - 2)5(x - 4)2.

The result will look like this: (x - 2)m(x - 4)n(<what's left>)

3. Dec 13, 2011

So it would be like this,
((x - 2)4(x - 4)2(5(x - 4) - 3(x - 2))) / (x - 4)6
( (x - 2)4(5(x - 4) - 3(x - 2)) ) / (x - 4)4

Now I stuck again.

Last edited: Dec 13, 2011
4. Dec 13, 2011

### Dick

You can simplify (5(x - 4) - 3(x - 2)) a bit more. Then if the derivative is going to zero, then the numerator has to be zero. Which values of x make that happen?