#### To determine

**To find:** The equation of the tangent line to the curve at the point.

#### Answer

The equation of tangent line to the curve y=11+x2 at (−1,12) is y=x2+1.

#### Explanation

**Given:**

The equation of the curve y=11+x2.

The curve passing through the point (−1,12).

**Derivative rules:**

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2x]2

(2) Power Rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

Here, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The derivative of y is dydx, which is obtained as follows.

dydx=ddx(y) =ddx(11+x2)

Apply the quotient rule (1) and simplify the terms,

dydx=(1+x2)ddx(1)−(1)ddx(1+x2)(1+x2)2

Apply the derivative rules (3) ,(2) and simplify the terms,

dydx=(1+x2)ddx(1)−(1)[ddx(1)+ddx(x2)](1+x2)2=(1+x2)(0)−(1)(0+2x2−1)(1+x2)2=0−2x(1+x2)2=−2x(1+x2)2

Therefore, the derivative of the function y=11+x2 is dydx=−2x(1+x2)2.

The slope of the tangent line at (−1,12) is,

m=dydx|x=−1=−2(−1)(1+(−1)2)2=24=12

Thus, the slope of the tangent line at (−1,12) is m=12.

Substitute (−1,12) for (x1,y1) and m=12 in the equation (1),

y−12=12(x−(−1))y−12=12(x+1)y=12x+12+12y=x2+1

Therefore, the equation of tangent line to the curve y=11+x2 at (−1,12) is y=x2+1.

#### To determine

**To sketch:** The graph of the curve and tangent line.

#### Explanation

**Given:**

The curve y=11+x2.

The tangent line y=x2+1 and the point is (−1,12).

**Graph:**

Use online graphing calculator and draw the graph of the curve and tangent line at (−1,12) is shown below in Figure 1.

From Figure 1, it is observed that the tangent line y=x2+1 touches the curve y=11+x2

at the point (−1,12).