SOLUTION: Consider the curve (a√1-bx) where a and b are constants. The tangent to this curve at the point where x=-1 is 3x+y=5. Find the values of a and b.

Algebra ->  Functions -> SOLUTION: Consider the curve (a√1-bx) where a and b are constants. The tangent to this curve at the point where x=-1 is 3x+y=5. Find the values of a and b.      Log On


   

Question 1192490: Consider the curve (a√1-bx) where a and b are constants. The tangent to this curve at the point where x=-1 is 3x+y=5. Find the values of a and b.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the tangent to the curve given by y=a%2Asqrt%281-bx%29=a%2A%281-bx%29%5E0.5 at the point with
is 3x%2By=5 , at the point x=-1 ,
the point of tangency is a point shared by the curve and the tangent line.
That point has x=-1 , and a y such that 3%28-1%29%2By=5 --> -3%2By=5 --> y=5%2B3 --> y=8 .
So, we know that the value of the function at x=-1 is
a%2Asqrt%281-b%28-1%29%29=8 <--> a%2Asqrt%281%2Bb%29=8 <--> highlight%28a%2A%281%2Bb%29%5E0.5=8%29
The slope of that tangent is the value of the derivative of the function at x=-1 .
That slope can be found by "solving" 3x%2By=5 for y to get
y=-3x%2B5 , so the slope is -3
The derivative is dy%2Fdx=-0.5ab%281-bx%29%5E%28-0.5%29=-ab%2F2sqrt%281-bx%29 ,
and for x=-1 the derivative value is -0.5ab%281%2Bb%29%5E%28-0.5%29=-3 <--> highlight%280.5ab%281%2Bb%29%5E%28-0.5%29=3%29 .
Dividing one highlighted equation by the other, we get
a%2A%281%2Bb%29%5E0.5%2F%280.5ab%281%2Bb%29%5E%28-0.5%29%29%22=%228%2F3 --> 2%281%2Bb%29%2Fb=8%2F3 --> 6%2B6b=8b --> 6=2b --> highlight%28b=3%29 .
Substituting the value found into
a%2Asqrt%281%2Bb%29=8 , we get a%2Asqrt%281%2B3%29=8 --> a%2Asqrt%284%29=8 --> 2a=8 --> highlight%28a=4%29 .
graph%28200%2C300%2C-5%2C5%2C-1%2C14%2C4%2Asqrt%281-3x%29%2C-3x%2B5%29