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

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SUMMARY

The discussion focuses on finding the y-intercept of the tangent line to a polynomial given the equation of a parallel line. The user outlines the process of determining the slope (m) from the derivative of the polynomial and the y-value from the polynomial itself. They express confusion about how to derive the y-intercept (b) without specific x or y values. Ultimately, the user realizes that they can express b as a function of x, clarifying their understanding of the tangent line's equation in slope-intercept form.

PREREQUISITES
  • Understanding of polynomial functions and their derivatives
  • Knowledge of tangent lines and their equations
  • Familiarity with point-slope and slope-intercept forms of linear equations
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the process of finding derivatives of polynomial functions
  • Learn how to derive the equation of a tangent line from a given point
  • Explore the concept of parallel lines and their slopes in calculus
  • Practice problems involving finding y-intercepts of tangent lines
USEFUL FOR

Students studying calculus, particularly those learning about derivatives and tangent lines, as well as educators looking for examples of common student misconceptions in this area.

J.J.T.
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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?
 
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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?
 
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.
 
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?
 
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)"?
 
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?
 
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.
 

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