College Algebra Exercises Section 1.1

As you study these exercises, move your cursor over arrows, equal symbols, light bulbs, and highlighted words. Always check in with the wise old owl and his little apprentice by moving your cursor over them. Study with an active cursor.

 
The definition that pops up when you move the cursor over a highlighted word is what should pop into your mind when you read, hear, or speak that word. The material that pops up when you move the cursor over a light bulb is a suggested strategy for solving the problem. You should always formulate a similar strategy when you attempt to solve a problem. When you move the cursor over an explanation of a step in a solution, the property that pops up is the mathematical justification for taking that action. You must always be able to provide such a justification for every step you take in mathematics.

9. Find the x-intercept of the graph of the equation y = 16 - 4x2

Solution: Let y = 0 to obtain 0 = 16 - 4x2 which we must solve for x.
16 - 4x2 = 0     divide both sides by 4
4 - x2 = 0       add - 4 to both sides
-x2 = -4          multiply both sides by -1
x2 = 4             take the square root of both sides
x = 2 or x = -2     substitute these into the original equation to determine if they are solutions
16 - 4(2)2 = 0    is a true statement so 2 is a solution
16 - 4(-2)2 = 0    is a true statement so - 2 is a solution
Therefore the x-intercepts of the graph of the equation y = 16 - 4x2 are 2 and -2.

49. Find the equation of the circle with center at (0, 0) and radius 4.

Solution: Substitute into the standard form x2 + y2 = r2 to get x2 + y2 = 16 as the desired equation.

51. Find the equation of the circle with center at (2, -1) and radius 4.

Solution: Substitute into the standard form (x - h)2 + (y - k)2 = r2 to get (x - 2)2 + (y + 1)2 = 16 as the desired equation.
This equation can be rewritten by performing the indicated squaring and then simplifying.
(x - 2)2 + (y + 1)2 = 16
x2 - 4x + 4 + y2 + 2y + 1 = 16
x2 + y2 - 4x + 2y = 11

54. Find the equation of the circle which contains the point (-1, 1) and center at (3, -2).

Solution: The radius of the desired circle is
Substitute this radius and the coordinates of the center into the general equation of a circle to obtain
(x - 3)2 + (y + 2)2 = 52
This equation can be rewritten by performing the indicated squaring and then simplifying.
(x - 3)2 + (y + 2)2 = 52
x2 - 6x + 9 + y2 + 4y + 4 = 25
x2 + y2 - 6x + 4y = 12

56. Find the equation of the circle which has a diameter with endpoints (-4, -1) and (4, 1).

Solution: The center of the circle is the point .
The radius is the distance from the origin (0, 0) to the point (4, 1) which is
Substitute this radius into the equation of a circle with center at the origin to obtain
x 2 + y 2 = 17    

58. Find the center and radius of the circle described by the equation x2 + y2 = 16. Sketch its graph.

Solution: If we rewite the given equation x2 + y2 = 16 as x2 + y2 = 42, it is obvious that it represents a circle with center at the origin (0, 0) with radius 4.
The graph is
:                    

62. Find the center and radius of the circle described by the equation (x - 2)2 + (y + 1)2 = 3. Sketch its graph.

Solution: If we rewite the given equation (x - 2)2 + (y + 1)2 = 3 as , it is obvious that it represents a circle with center at the origin (2, -1) with radius .
The graph is
: