the points p(1,-2) and q(2,k) lie on curve for which dy/dx=3-2/x. the tangents to the curve at the points p and q intersect at the point r

Answer 1

integrate the slope y' = dy/dx=3-2/x to get y= 3x -2lnx + c, the equation of the curve

substitute (1,-2) into that equation to get -2 = 3 + 0 + c or c=-2-3 (lnx=0) so c= -5

the equation of the curve containing points p and q is y=3x-2lnx -5 with slope y'=dy/dx

y'= the derivative or y'=3-2/x

substitute the point q(2,k) into the curve's equation to get: k=3(2)-2ln(2)-5=1-2ln2

k=1-2ln2

q(2,k) = q(2,1-2ln2) It's tangent line is (y-1+2ln2)/(x-2)=slope = 3-2/2)=3-1=2 using the point slope formula

or y=2(x-2)+1-2ln2 =2(x)-4-2ln2 =2x-4-2ln2

y=2x-4-2ln2

The tangent line through p(1-2) is (y+2)/(x-1)=1 or y=x-1-2 = x-3 using the point slope formula

y=x-3

subtract one tangent equation from the other to get: 0=x-1-2ln2

or x=1+2ln2 =2ln2+1

then y=x-3= 1+2ln2-3=2ln2-2

y=2ln2-2

The point of intersection is r(2ln2+1, 2ln2-2)

Here are the two tangent lines:

y=x-3 and y=2x-4-2ln2. they both contain the point r(2ln2+1, 2ln2-2)

substitute r into each tangent line equation. replace each x by 2ln2+1 and each y by 2ln2-2

2ln2-2=2ln2+1-3 and 2ln2 = 2(2ln2+1)-4-2ln2

=2ln2-2 =4ln2+2-4-2ln2

=2ln2-2

point r is on both tangent lines, where they intersect.

Answer 2

y = ∫(3 - 2/x)dx = 3x - 2lnlxl + C

Since (1,-2) is on the graph, 3 - 2ln(1) + C = -2. So, C = -5

y = 3x - 2lnlxl - 5

Since (2,k) is on the graph, k = 6 - 2ln2 - 5 = 1 - ln4

Find equations of the tangent lines:

Slope of tangent line at p: 3 - 2/1 = 1

Equation of tangent line at p: y + 2 = 1(x - 1)

y = x - 3

Slope of tangent line at q: 3 - 2/2 = 2

Equation of tangent line at q: y - (1 - ln4) = 2(x - 2)

y = 2x - 3 - ln4

Find r, the point of intersection of the tangent lines:

x - 3 = 2x - 3 - ln4

-x = -ln4

x = ln4

y = x - 3 = ln4 - 3

So, r = (ln4, ln4 - 3)