Finding equation to tangent line

• danielle36
In summary, the problem is finding the equation of the tangent line to the hyperbola y = 3/x at the point (3,1). Using the equation m = \frac{f(a+h) - f(a)}{h}, we can find the slope of the tangent line at that point. By simplifying the equation, we get m = \frac{-1}{3} as the slope. Similarly, in the second example, we are finding the slope of the tangent line to the parabola y = 4x - x^{2} at the point (1,3). Again, by simplifying the equation, we get m = 2 as the slope. This can be understood by taking the limit
danielle36
Hello,
I'm having trouble understanding an example from my test and I would appreciate your help clarifying how to get from one step to the next.

Problem: Find an equation of the tangent line to the hyberola y = 3/x at the point (3,1)
Eqation: $$m = \frac{f(a+h) - f(a)}{h}$$

$$m = \frac{f(3 + h) - f(3)}{h}$$

$$= \frac{\frac {3}{3 + h} - 1}{h}$$

$$= \frac{\frac {3 - (3 + h)}{3 + h}}{h}$$

(I'm assuming here they have just converted 1 to 3 + h / 3 + h, but I don't understand how they get to the next step...)

$$= \frac{-h}{h(3+ h)}$$

$$= \frac{-1}{3 + h}$$

$$= \frac{-1}{3}$$[/tex]

danielle36 said:
Hello,
I'm having trouble understanding an example from my test and I would appreciate your help clarifying how to get from one step to the next.

Problem: Find an equation of the tangent line to the hyberola y = 3/x at the point (3,1)
Eqation: $$m = \frac{f(a+h) - f(a)}{h}$$

$$m = \frac{f(3 + h) - f(3)}{h}$$

$$= \frac{\frac {3}{3 + h} - 1}{h}$$

$$= \frac{\frac {3 - (3 + h)}{3 + h}}{h}$$

(I'm assuming here they have just converted 1 to 3 + h / 3 + h, but I don't understand how they get to the next step...)
Yes,
$$\frac{\frac{3}{3+h}- 1}{h}=\frac{\frac{3}{3+h}- \frac{3+h}{3+h}}{h}= \frac{\frac{-h}{3+h}}{h}$$
And dividing by h is the same as multiplying by 1/h
$$\frac{\frac{-h}{3+h}}{h}= \frac{-h}{3+h}\frac{1}{h}= \frac{-h}{h(3+h)}$$

$$= \frac{-h}{h(3+ h)}$$

$$= \frac{-1}{3 + h}$$
By the way, this last equality is true as long as h is not equal to 0 since in that case we have 0/0. Fortunately, the limit, as h goes to 0 does not depend on the value at 0.

$$= \frac{-1}{3}$$

Ohh right! Thanks! That's what I figured was going on but I was't keeping the brackets in the denominator, I don't know why I couldn't see that now!

Ok, so I was going over this example again, and I've realized I'm not sure what is going on in that last step (where the last h is gone, seemingly at random)... Has 0 been substituted for h?

The limit as h goes to 0. That is not the same as "0 substituted for h", especially here because, as I said before "this last equality is true as long as h is not equal to 0 since in that case we have 0/0."
But I also said "Fortunately, the limit, as h goes to 0, does not depend on the value at 0". Since, as long as h is not o, this is the same as -1/(3+h) and that is continuous at h= 3, the limit can be calculated by setting h= 0 in that: -1/(3+ 0)= -1/3.

Hmm I'm not sure if I know what you mean, because I think I have fallen into a similar situation with a problem I was working on...

The solution in this case is 2..
Find the slop of the tangent line to the parabola y = 4x - x$$^{2}$$ at the point (1,3)

$$m = \frac{f(x+h) - f(x)}{h}$$
$$= \frac{(4[1+h] - [1 + h]^{2}) - 3)}{h}$$
$$= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}$$
$$= 2 - h$$

Then I'm not sure what to do with the h... so maybe this is the same situation? When can h not be set at 0?

Last edited:
By the time you are looking at the definition of "derivative" and trying to use it, you should already be familiar with "limits" and "continuity". Do you remember the definition of "[itex]\lim_{h\rightarrow 0} f(h)" and the condition under which it is equal to f(0)?

Yeah I'm familiar with the little limit symbol with the right arrow under it, obviously not familiar enough though... I do know where to find more information on it in my text though, thanks for showing me where to look!

danielle36 said:
Hmm I'm not sure if I know what you mean, because I think I have fallen into a similar situation with a problem I was working on...

The solution in this case is 2..
Find the slop of the tangent line to the parabola y = 4x - x$$^{2}$$ at the point (1,3)

$$m = \frac{f(x+h) - f(x)}{h}$$
$$= \frac{(4[1+h] - [1 + h]^{2}) - 3)}{h}$$
$$= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}$$
$$= 2 - h$$

Then I'm not sure what to do with the h... so maybe this is the same situation? When can h not be set at 0?
I wasn't asking if you were familiar with "the little symbol", but with the basic properties of limits and continuity!

$$= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}= \frac{2h- h^2}{h}$$
$$= \frac{h(2- h)}{h}$$
and, for h not equal to 0, the hs cancel out.
Now take the limit as h goes to 0. If h is "very, very, very" close to 0, then obviously 2- h is "very, very, very" close to 2- 0 and the limit is 2.

Recall that a function is said to be "continuous at a" precisely if [itex]\lim_{x\rightarrow a} f(x)= f(a). You should also be aware of theorems that tell you "all polynomials are continuous for all a", "all rational functions are continuous whenever the denominator is not 0". Taking the limit as h goes to 0 is NOT just "setting h= 0" but in those situations, or situations you can reduce to those, they are the same.

1. How do I find the equation to a tangent line?

To find the equation of a tangent line, you will need the coordinates of a point on the line and the slope of the line. You can use the slope formula or the derivative of the function to find the slope. Once you have the slope, you can use the point-slope formula to find the equation of the tangent line.

2. What is the point-slope formula?

The point-slope formula is used to find the equation of a line given a point on the line and the slope of the line. It is written as y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

3. Can I use the derivative to find the slope of the tangent line?

Yes, the derivative of a function at a given point represents the slope of the tangent line at that point. So, you can use the derivative to find the slope of the tangent line to a function at a specific point.

4. Is there a way to check if my equation is correct?

Yes, you can check your equation by plugging in the coordinates of the given point into the equation. The resulting y-value should match the y-coordinate of the given point. You can also graph the function and the tangent line to visually confirm if they intersect at the given point.

5. Can I find the equation of a tangent line to a curve that is not a straight line?

Yes, you can find the equation of a tangent line to a curve using the same methods described above. However, the equation may not be in a familiar form such as y = mx + b, as the tangent line may not be a straight line. Instead, the equation may be in terms of the function and its derivative.

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