#### To determine

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

#### Answer

The equation of the tangent line to the equation y2=x3+3x2 at the point (1,−2) is 9x+4y=1.

#### Explanation

**Given:**

The curve with equation y2=x3+3x2.

The point is (1,−2).

**Derivative rules:**

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable function, then

dydx=dydu⋅dudx.

(2) Product rule: If y=f(u) and u=g(x) are both differentiable function, then

ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x)).

**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,y=y1.

**Calculation:**

Obtain the equation of tangent line to the given point.

y2=x3+3x2

Differentiate the above equation implicitly with respect to *x*,

ddx(y2)=ddx(x3+3x2)ddx(y2)=ddx(x3)+3ddx(x2)ddx(y2)=3x2+6x

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

ddy(y2)dydx=3x2+6x2ydydx=3x2+6xdydx=3x2+6x2y

Therefore, the derivative of the equation is dydx=3x2+6x2y.

The slope of the tangent line at (1,−2) is computed as follows,

m=dydx|(x1,y1)=(1,−2)=3(1)2+6(1)2(−2)=−94

Thus, the slope of the tangent line at (1,−2) is m=−94.

Substitute (1,−2) for (x1,y1) and m=−94 in equation (1),

y+2=−94(x−1)y+2=−9x4+944y+84=−9x4+949x+4y=1

Therefore, the equation of the tangent line to the equation y2=x3+3x2 at the point (1,−2) is 9x+4y=1.

#### To determine

**To find:** The points if the curve has horizontal tangent.

#### Answer

The curve has horizontal tangent at (−2,−2) and (−2,2).

#### Explanation

**Given:**

The curve with equation y2=x3+3x2.

The derivative of the equation dydx=3x2+6x2y.

**Calculation:**

Obtain the point if the curve has horizontal tangent.

Note that, the curve horizontal tangent if dydx=0.

Suppose that dydx=0,

3x2+6x2y=03x2+6x=0x(3x+6)=0x=0 or x=−2

Rewrite the given equation as follows,

y=x3+3x2

Substitute x=0 in the above equation,

y=03+3(0)2=0

The derivative of the equation does not exist at (0,0).

Substitute x=−2 in y=x3+3x2,

y=(−2)3+3(−2)2=−8+12=4=±2

Therefore, the curve has horizontal tangent at (−2,−2) and (−2,2).

#### To determine

**To sketch:** The curve and tangent line to the given point.

#### Explanation

**Graph:**

Using online graphic calculator to draw the curve and the tangent line as shown below in Figure 1,

From Figure 1, it is observed that the line 9x+4y=1 touch the curve y2=x3+3x2 at (1,−2). That is, the tangent line 9x+4y=1 to the curve at (1,−2) and the curve has horizontal tangent at (−2,−2) and (−2,2).