Find f`(x) for the tangent line of the graph

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Homework Help Overview

The problem involves finding the derivative of a function at a specific point, as well as determining the equation of the tangent line to the graph of the function at that point. The context includes a tangent line that passes through two given points.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find the slope of the tangent line using the two points provided. There is uncertainty about the correct form of the tangent line equation. Some participants suggest using the general form of a line and checking the arithmetic involved in the calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the tangent line equation and checking the validity of the calculations. There is no explicit consensus on the correctness of the derived equation for the tangent line.

Contextual Notes

Participants note potential confusion regarding the interpretation of the equation and the arithmetic involved in verifying whether the points satisfy the equation of the tangent line. There is an emphasis on clarity in notation and calculations.

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Homework Statement


Suppose line tangent to graph of y=f(x) at x =3 passes through (-3, 7) & (2,-1).
Find f'(3), what is the equation of the tangent line to f at 3?


Homework Equations



I found the slope of which equals -8/5

Im not sure how to find the equation... do I do y-3=-8/5(x-3)? or is that wrong?
 
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The general form of a straight line is
y=m x + c
Try fitting that to your points
 
The tangent line passes through those two points.
Since it is a LINE, you may use the equation to solve for the slope:

m= (y2-y1) / (x2-x1)

Once you have found the slope, use any of of the two coordinates to solve for 'c', the constant/y-intercept.

Once you have the constant, simply write it in the form y=(#)x + (#).
That is the equation of the tangent line at that point x=3.
 
y=-8/5x+8.2 is this answer correct?
 
Why do you keep asking? It is simple arithmetic to check whether or not (-3, 7) and (2, -1) satisfy the equations you give.

Also, use parentheses to make your meaning clear. Many people would interpret "-8/5x" as "-8/(5x)". And I am not clear whether "8.2" means multiplication or "8 and 2 tenths". If you mean y= (-8/5)x+ (8)(2)= (-8/5)x+ 16 then if x= -3, y= (-8/5)(-3)+ 8.2= 24/5+ 8.2= 4. which is not 7. And if you mean y= (-8/5)x+ 8.2= (-8/5)x+ 82/10= (-8/5)x+ 41/5, then which x= -3, y= (-8/5)(-3)+ 41/5= 24/5+ 41/5= 65/5= 13, again, not 7.
 

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