How Do You Find the Equation of a Tangent Line to a Curve?

[teneleven]

Homework Statement

Find an equation of the tangent line to the curve at the given point.

$$y = \frac{x - 1}{x - 2}$$ , (3, 2)

Homework Equations

$$m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$$

$$y = mx + b$$

The Attempt at a Solution

$$\lim_{h\to 0} \frac{\frac{(x + h) - 1}{(x + h) - 2} - \frac{x - 1}{x - 2}}{h}$$

$$\lim_{h\to 0} \frac{\frac{2 + h}{1 + h} - \frac{2}{1}}{h}$$

$$\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2}{h}$$

$$\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2 + 2h}{h + h^2}$$

$$\lim_{h\to 0} \frac{h + 2h}{h + h^2}$$I'm not sure what to do after this step. The final answer is $$y = -x + 5$$
Thanks.

 

[arildno]

Your last line is wrong (remember to change + to -!).
We have:
$$\lim_{h\to{0}}(\frac{2+h}{h(1+h)}-\frac{2}{h})=\lim_{h\to{0}}\frac{2+h-2(1+h)}{h(1+h)}=\lim_{h\to{0}}\frac{-h}{h(1+h)}=\lim_{h\to{0}}\frac{-1}{(1+h)}=-1$$

 

[Architeuthis Dux]

Great job so far! After simplifying the limit, you can substitute the given point (3,2) into the equation y = mx + b to solve for the value of b. This will give you the equation of the tangent line at (3,2). Here's how it would look:

y = \frac{x - 1}{x - 2} , (3, 2)

m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}

m = \lim_{h\to 0} \frac{\frac{(3 + h) - 1}{(3 + h) - 2} - \frac{3 - 1}{3 - 2}}{h}

m = \lim_{h\to 0} \frac{\frac{2 + h}{1 + h} - \frac{2}{1}}{h}

m = \lim_{h\to 0} \frac{h + 2h}{h + h^2}

m = \lim_{h\to 0} \frac{3h}{h + h^2}

m = \frac{3}{1 + 0} = 3

Now, substituting the given point (3,2) into y = mx + b, we get:

2 = 3(3) + b

2 = 9 + b

b = -7

Therefore, the equation of the tangent line at (3,2) is y = 3x - 7. Great job!