SOLUTION: Let f(x)=x^3-x^2-5x-1. Find an equation of the line tangent to the graoh of f at the point (2,f(2)). Find an equation of the line tangent to the graph of f(x)= (x^3+1)/√x

Algebra ->  Graphs -> SOLUTION: Let f(x)=x^3-x^2-5x-1. Find an equation of the line tangent to the graoh of f at the point (2,f(2)). Find an equation of the line tangent to the graph of f(x)= (x^3+1)/√x      Log On


   

Question 1008240: Let f(x)=x^3-x^2-5x-1. Find an equation of the line tangent to the graoh of f at the point (2,f(2)).
Find an equation of the line tangent to the graph of f(x)= (x^3+1)/√x at the point corresponding to x=1.
please.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The value of the slope of the line is equal to the value of the derivative at that point.
f%28x%29=x%5E3-x%5E2-5x-1
df%2Fdx=3x%5E2-2x-5
So at x=2,
m=3%282%29%5E2-2%282%29-5
m=12-4-5
m=3
The value of the function is,
f%282%29=2%5E3-2%5E2-5%282%29-1
f%282%29=8-4-5-1
f%282%29=-6
So now use the point-slope form of a line,
y-%28-6%29=-4%28x-1%29
y%2B6=-4x%2B4
highlight%28y=-4x-2%29
.
.
.
.
.
.
.
Similarly,
y=%28x%5E3%2B1%29x%5E%28-1%2F2%29=x%5E%285%2F2%29%2Bx%5E%28-1%2F2%29
dy%2Fdx=%285%2F2%29x%5E%283%2F2%29-%281%2F2%29x%5E%28-3%2F2%29
So at x=1
m=%285%2F2%29%281%29%5E%283%2F2%29-%281%2F2%29%281%29%5E%28-3%2F2%29
m=5%2F2-1%2F2
m=2
and
y%281%29=%281%2B1%29%2Fsqrt%281%29=2
So point slope form is,
y-2=2%28x-1%29
y-2=2x-2
highlight%28y=2x%29
.
.
.
.