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).