practice5 sec 2.8, 3.1, 3.2

1]

Solve and graph. Express answer in both inequality and interval notation.

SOLUTION

Solve the equation

to determines the zero of the function.

The only zero is

.

We must exclude the points where the denominator is equal to zero,
because we can not divide by zero.
The solutions of the equation

are

.

let's indicate the points corresponding to these numbers on the number line.
To facilitate talking about the fraction, let

.

Evaluate

at a single point in each of the subintervals defined by the points -2, 0, and 4.

is positive at single point, then it is positive at every point in the subinterval.
Likewise, if it is negative at a single point, then it is negative at all the points in the subinterval.
The above graph is the result of such computations.
For example,

From the graph we can read the solution set for the problem.
We are looking for those intervals with the negative signs.
The solution set is

     0 and -2 are excluded from the solution set because at those points
     the denominator is zero and division by zero is inadmissible.
     4 is included in the solution set because the inequality in the problem is

     "

"

The rational function can be equal to zero.

2]
     Find the center and the radius of the circle with the given equation.
      Graph the equation.

SOLUTION

We need to put the equation in the standard form of the equation of a circle

in order to read the coordinates of the center

and the radius

.

Completing the squares on the

and the

variables, we get the equation

.

So the center of the circle is at

and the radius is

.

3]
     Write an equation for the line that contains the indicated point
     and satisfies the indicated condition.

     Write the answer in the standard form

.

(a)

, parallel to

.

SOLUTION

is in the slope-intercept form

,

so the slope of the given line is 4
and the slope of any other line parallel to it is also 4.
We use the point-slope form of the equation of the straight line

to finish the problem.
Substitute the coordinates of the given point

and we get

   are equivalent answers to the problem.

     (b)

, perpendicular to

SOLUTION
Writing the given line in the slope-intercept form

,

we see that the slope

.

So the slope of any other line perpendicular to the given line is

.

The given point is

.

As in the previous problem:


		Algebra B Practice 5 Module 2 & 3 Problems relating to sec 2.8, 3.1, 3.2

	1] Solve and graph. Express answer in both inequality and interval notation. SOLUTION Solve the equation to determines the zero of the function. The only zero is . We must exclude the points where the denominator is equal to zero, because we can not divide by zero. The solutions of the equation are . let's indicate the points corresponding to these numbers on the number line. To facilitate talking about the fraction, let . Evaluate at a single point in each of the subintervals defined by the points -2, 0, and 4. If is positive at single point, then it is positive at every point in the subinterval. Likewise, if it is negative at a single point, then it is negative at all the points in the subinterval. The above graph is the result of such computations. For example, From the graph we can read the solution set for the problem. We are looking for those intervals with the negative signs. The solution set is 0 and -2 are excluded from the solution set because at those points the denominator is zero and division by zero is inadmissible. 4 is included in the solution set because the inequality in the problem is " " The rational function can be equal to zero. 2] Find the center and the radius of the circle with the given equation. Graph the equation. SOLUTION We need to put the equation in the standard form of the equation of a circle in order to read the coordinates of the center and the radius . Completing the squares on the and the variables, we get the equation . So the center of the circle is at and the radius is . 3] Write an equation for the line that contains the indicated point and satisfies the indicated condition. Write the answer in the standard form . (a) , parallel to . SOLUTION is in the slope-intercept form , so the slope of the given line is 4 and the slope of any other line parallel to it is also 4. We use the point-slope form of the equation of the straight line to finish the problem. Substitute the coordinates of the given point and we get or are equivalent answers to the problem. (b) , perpendicular to SOLUTION Writing the given line in the slope-intercept form , we see that the slope . So the slope of any other line perpendicular to the given line is or . The given point is . As in the previous problem: or are equivalent answers to the problem. top © edmond 2003