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

#1**+2 **

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

#2

#3**+1 **

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**+1 **

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**+1 **

Yep! No need to re-do it, I copied your steps with the correct derivative. Thanks :)

Julius
Feb 19, 2018

#6**+2 **

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

ElectricPavlov
Feb 19, 2018