f(x)=lnx
y=f(x)
dy/dx= 1/x
tangent at (x,y) has slope 1/x
eqn of tangent is y = mx + c
since the tangent passes through origin, c=0
substitute y = lnx and m= 1/x to above eqn
lnx = 1
x=e
y=lne=1
Answer:
The point slope formula is
(y−y1)=m(x−x1)
Where m = the slope calculated as y2−y1x2−x1
(x1,y1)(x2,y2)
(-4 , 1) (-2 , 3)
Solve for the slope
m = 3−1−2−(−3) = 21 = m = 2
(y−3)=2(x−(−2)) Plug in known values.
y−3=2(x+2) Simplify signs
y−3=2x+4 Use Distributive property
y=−2x+4+3 isolate the y variable
answer (y = −5x+7 ) Simplify
(X)(3.25)(4)= 110.5
4
(x)(3.25)= 27.625
3.25
x = 8.5
(8.5)(3.25)(4)= 110.5
