# At what point on the graph of y=1+(1/x) will the tangent line pass through the point (3,0)?

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Let the point be (x,y). Now the slope of the tangent to a point on any curve is the first derivative of the curve.

Here y = 1 + 1/x

=> y' = -1/x^2

Now, the slope of a line between (x1, y1) and (x2, y2) is (y2-y1)/ (x2-x1)

So (0-y)/ 3-x) = -1/x^2

=> y/(3-x) = 1/x^2

=> (1 + 1/x)/(3-x) = 1/x^2

=> (x+1)/(3-x) = 1/x

=> x*(x+1)= 3-x

=> x^2 + x = 3-x

=> x^2 + 2x - 3 =0

=> x^2 + 3x - x - 3 =0

=> x(x+3) -1(x+3) = 0

=> (x-1)(x+3) = 0

x= 1 or -3.

y= 1 + 1/x = 1+1 =2

or 1 + 1/x = 1 - 1/3 = 2/3

**Therefore the required point is (1, 2) and (-3 , 2/3)**.

The tagent to the curve y = 1+1/x is given by .

y-y1 = (dy/dx) (x-x1).

So the slope of the tangent dy/dx is gotten by differentiating y = 1+1/x.

dy/dx = (1+1/x)' = -1/x^2.

Therefore at x1 dy/dx = -1/x1^2. At x1, y1 = 1+1/x1.

So the tangent (at x1,y1) is given by:

y- (1+1/x1) = (-1/x2^2 (x-x1).....(1) Since this line passes through (3,0) , it should satisfy eq (1).

0-1-1/x1 = (-1/x1^2) (3-x1)

(-1-1/x1)(-x1^2) = 3-x1.

x1^2+x1+x1-3 = 0.

x^2 +2x-3 = 0.

(x1+3)(x1-1).

So x1 = -3 , or x1 = 1.

Therefore y1 = 1+1/x1 . Or y1 = 1-1/3 = 2/3 when x= -3

y1 = 1+1/1 = 2 , when x1 = 1.

Therefore the tangents at (-3,2/3) and (1, 2) on the curve y = 1+1/x passes the point (3,0).

The tangent line is the derivative of the given function:

dy/dx = -1/x^2

The slope of the tangent line is:

m = -1/3^2

m = -1/9

The equation of the line that passes through the point (3,0) and it has the slope m = -1/9 is:

y - 0 = (-1/9)(x - 3)

**y = -x/9 + 1/3**

On the other hand, the tangent line is passing through the point (3, 0) and the slope of a line that passes through 2 points is:

m = (y - 0)/(x - 3)

m = y/(x - 3)

We'll put m = dy/dx:

y/(x - 3) = -1/x^2

We'll cross multiply and we'll get:

x^2*y = 3 - x

x^2(1 + 1/x) = 3 - x

x^2 + x + x - 3 = 0

x^2 + 2x - 3 = 0

We'll can write a quadratic if we know the sum and the product of the roots:

x^2 - Sx + P = 0

S = -2

P = -3

x1 = -3 and x2 = 1

-3+1 = -2 = S

-3*1 = -3 = P

The quadratic could be written as a product of linear factors:

(x-1)(x+3) = 0

x1 - 1 = 0

**x1 = 1**

x2 + 3 = 0

**x2 = -3**

y1 = 1 + 1/x1

y1 = 1 + 1

**y1 = 2**

y2 = 1 - 1/3

**y2 = 2/3 **