Equation of tangent line to curve

[hadizainud]

Homework Statement

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

Homework Equations

Quotient Rule - Differentiation

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.

 

[Mark44]

Homework Statement

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

Homework Equations

Quotient Rule - Differentiation

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.

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>)

 

[hadizainud]

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.

 

[Dick]

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.

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?