#### To determine

**To find:**

The number of tangent lines to the curve

y=xx+1

pass through the point (1, 2) and at which point these lines touch the curve.

#### Answer

There are 2 tangent lines to the curve at the point (1, 2) and the point of contact are

(-0.27, -0.37) and (-3.73, 1.37)

#### Explanation

**1) Formula:**

i. Quotient rule:

ddxfxgx=gxddxfx-fxddx( gx)gx2

ii. Sum rule:

ddxfx+gx=ddxfx+ddx(gx)

iii. Constant function rule:

ddxC=0

iv. Quadratic formula:

x=-b± b2-4ac2a

v. Point-slope formula:

y-y1=m(x-x1)

**2) Given: **Equation of curve is,

y=xx+1

**3) Calculation:**

Consider the equation,

y=xx+1

The function is continuous at every value of x except at x = -1

Differentiate with respect to x,

dydx=ddx(xx+1)

By using quotient rule,

dydx=x+1ddxx-x ddx(x+1)x+12

By using sum rule,

dydx=x+1ddxx-x [ddx(x)+ddx(1)]x+12

By using constant function rule,

dydx=x+1 -x [1+0]x+12

dydx=x+1-xx+12

⇒dydx=1x+12

Consider a point at x = a on the curve. Then we know that slope of tangent at this point is

dydxa=1a+12.

By substituting a in equation of the curve we see that the point under consideration is a,aa+1

Therefore, by point-slope formula the equation of the tangent is

y-aa+1=1a+12(x-a)

This is the equation of a tangent through an arbitrary point a,aa+1 on the curve

Suppose the tangent at some point a,aa+1 is passing through (1, 2) then we have

2-aa+1=1a+12(1-a)

⇒2a+2-aa+1=1-aa+12

⇒a+2a+12a+1=1-a

⇒a+2(a+1)=1-a

⇒a2+3a+2=1-a

⇒a2+3a+2-1+a=0

⇒a2+4a+1=0

By using quadratic formula,

a=-4±42-4(1)(1)2(1)=-4±16-42=-4±122=-4±232=-2±3

It has 2 solutions hence there are 2 tangent lines to the curve passing through point (1, 2)

To find the points of tangency of these tangents we need to solve for y-coordinates Now y coordinate is,

For a=-2+3= -0.27

y=aa+1=-2+3-2+3+1=-2+3-1+3=-0.37

For a=-2-3= -3.73

y=aa+1=-2-3-2-3+1=-2-3-1-3=1.37

Therefore, the points at which the tangent lines through (1,2) touches the curve are

(-0.27, -0.37) and (-3.73, 1.37)

**Conclusion:**

There are 2 tangent lines to the curve passing through the point (1, 2) and the points of contact are (-0.27, -0.37) and (-3.73, 1.37).