
#1
Feb2612, 07:27 PM

P: 75

1. The problem statement, all variables and given/known data
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? 2. Relevant equations I found the slope of which equals 8/5 Im not sure how to find the equation... do I do y3=8/5(x3)? or is that wrong? 



#2
Feb2612, 07:40 PM

P: 615

The general form of a straight line is
y=m x + c Try fitting that to your points 



#3
Feb2612, 08:25 PM

P: 8

The tangent line passes through those two points.
Since it is a LINE, you may use the equation to solve for the slope: m= (y2y1) / (x2x1) Once you have found the slope, use any of of the two coordinates to solve for 'c', the constant/yintercept. 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. 



#4
Feb2612, 10:17 PM

P: 75

find f`(x) for the tangent line of the graph
y=8/5x+8.2 is this answer correct?




#5
Feb2712, 07:56 AM

Math
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P: 38,896

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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