Calc 1 practice exam question -- finding the y intercept of the tangent line

In summary: You are given the equation for a line that is parallel to the tangent line. You are asked to find the y-intercept of the line.
  • #1
J.J.T.
17
6
I was given the equation of a polynomial told to find the derivative. easy enough.

Then asked to give the equation of the tangent line which I've only learned how to get in the form of the question: "find the equation of the tangent line at x="
They gave me the equation of a line parallel to the tangent.

Typically as we've seen these problems thus far its fairly straightforward derive equation for slope of tangent plug x into the original equation to find the y value and typically put in point-slope form. Alternatively i suppose you could find the y-intercept and put in y=mx+b form but,

How in the heck do you find either the y intercept of the tangent line or the x/y points to do that given only the equation of a parallel line?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You have the slope of the polynomial at x. That's m.

You have the value of y at x. That's mx+b.

You have x.

So in y=mx+b you have y, m, and x. That let's you solve for b. Right?

Now you have y=mx+b for the tangent line.

So what is the y intercept for this line?
 
  • #3
No we weren't given the "x" value or "y" value at which we should take the tangent. But were asked to write the equation of the tangent line. It seems like they only really gave us two ways to find the slope.
 
  • #4
Ok, I will hold your hand a little more.

You have the values of x, y, and m as functions of x. The value of x as a function of x is just x. :wink:

The value of y as a function of x is the polynomial.

The value of m as a function of x is the slope of the polynomial.

So you can work out y = mx + b, and solve for b as a function of x. Because you have y, m, and x, as functions of x.

Right?
 
  • #5
Even if i did that it would leave b still in the form of a function of "x" without knowing the y or x-value how am i supposed to write the equation in a familiar form e.g. point-slope or slope-intercept?
The way you're describing it I would end up with an equation that looks like "y=(derivative of polynomial)x+(b as a function of x)"?
 
  • #6
J.J.T. said:
Even if i did that it would leave b still in the form of a function of "x" without knowing the y or x-value how am i supposed to write the equation in a familiar form e.g. point-slope or slope-intercept?
At the y-intercept for a line, what is the value of x?
 
  • #7
Oh dang thanks for clearing that up. Wow... can't believe i missed that. It just threw me off my game because it was different from any of the hw/quiz problems.
 

1. What is the purpose of finding the y-intercept of the tangent line?

The y-intercept of the tangent line represents the point where the tangent line intersects the y-axis. It is useful in determining the initial value or starting point of a function at a specific point.

2. How do I find the y-intercept of a tangent line?

To find the y-intercept of a tangent line, you first need to find the slope of the tangent line at a given point. This can be done using the derivative of the function at that point. Once you have the slope, you can use the point-slope form of a line to find the equation of the tangent line. The y-intercept will be the value of y when x is equal to 0.

3. Can you provide an example of finding the y-intercept of a tangent line?

Sure, let's say we have the function f(x) = x^2 at the point (2,4). To find the y-intercept of the tangent line, we first find the derivative of the function, which is f'(x) = 2x. Then, we plug in the x-value of the given point into the derivative, giving us a slope of 4. Using the point-slope form with the given point (2,4), we get the equation of the tangent line as y = 4x - 4. Therefore, the y-intercept of the tangent line is -4.

4. What is the significance of the y-intercept of the tangent line?

The y-intercept of the tangent line can provide valuable information about the behavior of a function at a given point. It can help determine the direction of the function, whether it is increasing or decreasing, and can also give insight into the concavity of the function.

5. How can I use the y-intercept of the tangent line in real-life applications?

The y-intercept of the tangent line can be used in many real-life applications, such as in physics to determine the position of an object at a specific time or in economics to estimate the starting point of a company's growth. It can also be used in engineering to determine the initial value of a system at a certain time.

Similar threads

Replies
5
Views
1K
Replies
1
Views
943
Replies
1
Views
1K
Replies
4
Views
2K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
4
Views
1K
Replies
5
Views
2K
  • Calculus
Replies
4
Views
2K
Back
Top