SOLUTION: a quadratic function is defined by f(x)= 3x^2+4x-2. A linear function is defined by g(x)=mx-5. What value(s) of the slope owuld make it a tangeant to the prabola The answer

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: a quadratic function is defined by f(x)= 3x^2+4x-2. A linear function is defined by g(x)=mx-5. What value(s) of the slope owuld make it a tangeant to the prabola The answer      Log On


   

Question 246206: a quadratic function is defined by f(x)= 3x^2+4x-2. A linear function is defined by g(x)=mx-5. What value(s) of the slope owuld make it a tangeant to the prabola

The answer to the problem is m=-2, and m=10
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Is this question part of a calculus course? The answer
is easiest with calculus.
f%28x%29+=+3x%5E2+%2B+4x+-+2
fprime%28x%29+=+6x+%2B+4
g%28x%29+=mx+-+5
gprime%28x%29+=+m
I want to know all solutions x%5B1%5D%2Cy%5B1%5D
where
fprime%28x%5B1%5D%29+=+gprime%28x%5B1%5D%29
6x+%2B+4+=+m
Also, x%5B1%5D%2Cy%5B1%5D is a solution for
f%28x%29+=+g%28x%29
3x%5E2+%2B+4x+-+2+=+mx+-+5
3x%5E2+%2B+4x+-+2+=+%286x+%2B+4%29%2Ax+-+5
3x%5E2+%2B+4x+-+2+=+6x%5E2+%2B+4x+-+5
3x%5E2+=+3
x%5E2+=+1
x+=+1
x+=+-1
and
fprime%281%29+=+6%2A1+%2B+4
fprime%281%29+=+10
gprime%281%29+=+m
since
fprime%281%29+=+gprime%281%29
m+=+10
fprime%28-1%29+=+6%2A%28-1%29+%2B+4
fprime%28-1%29+=+-2
gprime%28-1%29+=+m
m+=+-2
I'll do a plot of the quadratic and 2 lines:
+graph%28+600%2C+600%2C+-4%2C+4%2C+-7%2C+8%2C+3x%5E2+%2B+4x+-+2%2C+10x+-+5%2C-2x+-+5+%29+