1. f(x)=x³-2x² at the point where x=1.

 

 
 

Solution:  f ^¢(x)=3x²-4x.  f ^¢(1)=-1.  Also, f(1)=-1.

Therefore, the slope of the line is -1, and a point on it is (1,-1).

The equation of the line is y-1=-1(x-1), which simplifies to y=-x+2.
 
 
 

2. g(x)=[x/(x+4)] at the origin.

 

 
 

Solution:  g ^¢(x)=[4/((x+4)²)].  g ^¢(0)=[1/4].

Therefore, the slope of the line is [1/4], and (0,0) is on it.

The equation of the line is y=[1/4]x.
 
 
 

3. h(x)=cos(x²) at the point where x=Ö{[(p)/6]}.

 

 
 

Solution:  h ^¢(x)=-2xsin(x²).  h ^¢(Ö{[(p)/6]})=-Ö{[(p)/6]}.  Also, h(Ö{[(p)/6]})=[(Ö3)/2].

Therefore, the slope of the line is -Ö{[(p)/6]}, and point is (Ö{[(p)/6]},[(Ö3)/2]).

The equation of the line is y-[(Ö3)/2]=-Ö{[(p)/6]}(x-Ö{[(p)/6]}), which simplifies to y=-Ö{[(p)/6]}x+Ö{[(p)/6]}+[(Ö3)/2].
 
 
 

1. f(x)=6-x².

 

 


 

Domain: (-¥,¥)

Range: (-¥,6]
 
 
 

2. xy=1.

 

 


 

Domain: (-¥,0)È(0,¥)

Range: (-¥,0)È(0,¥)
 
 
 

3. 4x²+9y²-16x+18y=11

 

 
 

This is the equation of a conic section.  It will be easier to graph if we complete the squares first.
 
 

4x2+9y2-16x+18y  = 11  4x2-16x+9y2+18y  = 11  4(x2-4x)+9(y2+2y)  = 11  4(x2-4x+4-4)+9(y2+2y+1-1)  = 11  4(x2-4x+4)-16+9(y2+2y+1)-9  = 11  4(x2-4x+4)+9(y2+2y+1)  = 36  4(x-2)2+9(y+1)2 = 36  (x-2)2 9 + (y+1)2 4 = 1

So, this is an ellipse centered at (2,-1).



Domain is [-1,5]

Range is [-3,1]
 
 
 

4. 4x²-9y²-16x-18y=29.

 

 
 

Again, we complete the square.

4x2-9y2-16x+18y  = 29  4x2-16x-9y2+18y  = 29  4(x2-4x)-9(y2-2y)  = 29  4(x2-4x+4-4)-9(y2-2y+1-1)  = 29  4(x2-4x+4)-16-9(y2-2y+1)+9  = 29  4(x2-4x+4)-9(y2-2y+1)  = 36  4(x-2)2-9(y-1)2 = 36  (x-2)2 9 - (y-1)2 4 = 1

So this is a hyperbola centered at (2,1).



Domain is (-¥,-1]È[5,¥)

Range is (-¥,¥)
 
 
 

5. The parametric curve described by x=2t+4,y=t-1.

 

 
 

Let's eliminate the parameter t.

x  = 2t+4  t  = x-4 2 y  = æ ç è x-4 2 ö ÷ ø -1  y  = 1 2 x-3

So this is a straight line.  Because there is no constraint on the parameter t, this is the complete line, not just a segment or ray.



Domain is (-¥,¥)

Range is (-¥,¥)
 
 
 

6. The parametric curve described by x=2+cost,y=1+3sint.

 

 
 

Again, we eliminate the parameter.

( cost) 2+( sint) 2 = 1  (x-2)2+ æ ç è y-1 3 ö ÷ ø 2     = 1  (x-2)2+ ( y-1) 2 9 = 1

So this is an ellipse centered at (2,1).



Domain is [1,3]

Range is [-2,4]
 
 
 

The area is given by the definite integral ò_x=-1^x=23x²+2   dx

ó õ x=2 x=-1 3x2+2   dx  = x3+2x]x=-1x=2 = 10-(-3)  = 13 square units

Suppose (x,y) is a point on the line.  Then the distance from that point to (2,3) is given by the following formula:       d=Ö{(x-2)²+(y-3)²}.

Because the point is on the line, y=2x, the distance can be written as d=Ö{(x-2)²+(2x-3)²}.  Now, we make things easier by realizing that if we find the x-value that minimizes the distance, we will also have the value that minimizes the square of the distance.  The square of the distance is given by d²=(x-2)²+(2x-3)².  This is the function we want to minimize.  Recall that we do this by setting its derivative equal to 0.

f(x)  = (x-2)2+(2x-3)2 f(x)  = x2-4x+4+4x2-12x+9  = 5x2-16x+13  f ¢(x)  = 10x-16

Setting 10x-16=0, gives x=8/5.  Since the function is a parabola that opens upward, this x-value must be a minimum point, not a maximum point.  The distance that corresponds to this x-value is Ö{([8/5]-2)²+(2([8/5])-3)²}=Ö{([(-2)/5])²+([1/5])²}=Ö{[4/25]+[1/25]}=Ö{[1/5]} units.