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# Tangent

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Consider the function f(x) = 5x2 + 3.  Determine the point at which the tangent to the curve f(x) at the point x = -2 intersects the function g(x) = x.

I got no idea for this one :|

Julius  Feb 19, 2018
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#1
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At x = -2    f(x) = 5 (-2^2) + 3 = 23        y = 23

The slope of f(x) is the derivative      10x    and at -2   this equals  -20

So we have a point  -2,23   and slope = -20

y=mx+b form of this line    23 = -20(-2) + b      or b = -17

so the tangent line is    y = -20x-17   Now just set this equal to g(x)

g(x) = x      x = -20x-17       21x=-17     x =  -17/21 = -.81

So the point of intersection is    -.81 , -..81

Here is a graph:

ElectricPavlov  Feb 19, 2018
edited by ElectricPavlov  Feb 19, 2018
#2
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Isn't the derivative of f(x)=5x^2+3 = f'(x)=10x

Julius  Feb 19, 2018
#3
+714
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So the Point of Intersection would be: (-17/21,-17/21)

Since the actual slope is -20

23=-20(-2)+b

b=-17

y=-20x-17

Setting them equal to one another

x=-17/21 -- Plugging in to find y

POI = (-17/21,-17/21)Is that correct?

Julius  Feb 19, 2018
#4
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You are correct Julius   ...I got the derivative of  5x^2 + 3x.....

sorry the solution methodology is the same though ...I will correct it.

ElectricPavlov  Feb 19, 2018
#5
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Yep! No need to re-do it, I copied your steps with the correct derivative. Thanks :)

Julius  Feb 19, 2018
#6
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...and you got the correct answer!   Good job, Julius.  Thanx for finding my error...I corrected it anyway.

ElectricPavlov  Feb 19, 2018
#7
+714
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Nope, thank you for all the help :)

Julius  Feb 19, 2018

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