SOLUTION: I need some help: a) for the following function and values of a, find f '(a) b) Determine an equation of the line tangent to the graph of f at the point (a, f(a)) for the given v

Algebra ->  Finance -> SOLUTION: I need some help: a) for the following function and values of a, find f '(a) b) Determine an equation of the line tangent to the graph of f at the point (a, f(a)) for the given v      Log On


   



Question 1087681: I need some help:
a) for the following function and values of a, find f '(a)
b) Determine an equation of the line tangent to the graph of f at the point (a, f(a)) for the given value of a.
F(x) = 4x%5E2+%2B+2x; a = -2

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
They want the derivative of F(x) at the point
( a, F(a) ) when +a+=+-2+, so the point is
( -2, F(-2) )
and F(-2) = +4%2A%28-2%29%5E2+%2B+2%2A%28-2%29+
F(-2) = +16+-+4+
F(-2) = +12+
---------------------
F'(x) = +8x+%2B+2+
F'(-2) = +8%2A%28-2%29+%2B+2+
F'(-2) = +-16+%2B+2+
F'(-2) = +-14+
What you now have is the slope of the line
tangent to F(x) at the point ( -2, 12 )
-----------------------------------------
I can call this line +G%28x%29+ and it looks like:
+G%28x%29+=+-14x+%2B+b+
Now use the given point ( -2, 12 )
+12+=+-14%2A%28-2%29+%2B+b+
+b+=+12+-+28+
+b+=+-16+
+G%28x%29+=+-14x+-+16+ is the equation of the tangent
-----------------------------
check:
Here's the plots:
+graph%28+400%2C+400%2C+-6%2C+6%2C+-5%2C+50%2C+4x%5E2+%2B+2x%2C+-14x+-+16+%29+
Looks OK