+0  
 
+1
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avatar+946 

 

 

I asked a question like this about perpendicular lines, and I just wanted to clarify if I did this right following the same procedures for the perpendicular line. 

 

f'(x) = 6x+1 

6x+1=4 

x = 2 

f(2) = 14 

(2,14) 

 

Is that correct?

Julius  Mar 5, 2018
 #1
avatar+88980 
+2

You have the correct idea....but a lttle mistake on the math

 

Derivative   =   6x  + 1

 

Set  =   to the slope of the line

 

6x + 1   =  4      

6x  = 3

x  =  3/6  = 1/2

 

And   y  =   3(1/2)^2  + 1/2   =  5/4

 

So.....the point  is   ( 1/2, 5/4)    and the equation of the line is

 

y  = 4(x - 1/2)  + 5/4

y = 4x - 2 + 5/4

y  = 4x - 3/4

 

Here's the graph, Julius : https://www.desmos.com/calculator/48aijjgr6d

 

 

cool cool cool

CPhill  Mar 5, 2018
 #2
avatar+946 
+1

Oh, my! How could I make such a mistake surprise

 

Thanks, CPhill! 

Julius  Mar 5, 2018
 #3
avatar+88980 
+1

No big deal....everyone makes mistakes.....

 

 

cool cool cool

CPhill  Mar 5, 2018

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