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# Perpendicular Lines?

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Okay so I found the derivative of f(x) which is f'(x) = -2x+3. Then I set it equal to the function y and found the point of intersection. So that was the answer for a as I'm guessing

Then for b I used that Point of intersection and the perpendicular slope of 1/3x+2 which is -3 and found the equation of the line. I feel I'm doing something wrong though...

Mar 3, 2018

#1
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Hopefully Chris is doing the math and comes up with the same answer I calculated...here is my graph

Mar 3, 2018
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Mar 3, 2018
edited by CPhill  Mar 3, 2018
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Ok    -2x +3   gives the SLOPE of the parabola at any point x

We want the slope to be -3   (negative inverse of 1/3.....the slope of the line)

so   -2x+3 = -3    yields  x = 2 is where the tangent line is slope -3

Sustitute x= 2 into the ORIGINAL equation to find y = 2

Now find the line  given slope = -3   and point  3,2

y= mx+b

2= -3(3  ) + b       so b = 11

y = -3x + 11      (see graph)

Mar 3, 2018
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Nice, EP  !!!!

CPhill  Mar 3, 2018
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Thanx, CP.......once in a while I don't make an error!   Haha

ElectricPavlov  Mar 3, 2018
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I'm a bit confused on how there is a point 3,2. :|

Mar 4, 2018
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The question is...

What are the coordinates of the point on the graph of  f(x)  where the slope of f(x)  =  -3  ?

the slope of f(x) at any  x  value  =  f'(x)  =  -2x + 3

What  x  value makes the slope of f(x) be  -3  ?

What  x  value makes  f'(x)  be  -3 ?

f'(x)  =  -3

-2x + 3  =  -3

-2x  =  -6

x  =  3

When  x = 3 ,  the slope of f(x)  =  -3

When  x = 3 ,  f(x)  =   f(3)  =  2

So at the point  (3, 2) , the slope of  f(x)  is  -3 .

Mar 4, 2018
edited by hectictar  Mar 4, 2018