Lines, Circles, and Piecewise Functions - College Algebra

The slope of a line determines its steepness.

Since slope is rise over run or

,

each of the slopes can be compared based on the ratio of rise to run.

A greater rise than run means a steeper line, or in our case, a hill. So for the "steepest" line , one must rise 5 units and move horizontally 1 unit.

Compare this with the line where one must rise one unit and move 8 units horizontally.

The standard form of a circle is the following:

where r is the radius and (a, b) is the center. In order to rework what we have into this form, we have to complete the square.

Since

we need to find an a so that 2a=4.

.

For the y coordinate, we need to find the same thing except instead we have the equation 2b=6.

.

And then we have the coordinates of the circle's center:

The -intercept of the graph of occurs at the point at which - that is, at the point with coordinates . By definition, if , so we want to find this value. To do this, substitute accordingly:

Solve for by isolating it on the left; first, subtract 15:

Divide by 7:

The -intercept of the graph of is at .

The -intercept of the graph of a function is the point at which it intersects the -axis. The -coordinate is 0, so the -coordinate can be found by evaluating .

The function is piecewise-defined, so it is necessary to use the definition applicable for . However, the first definition applies for values of less than 0, and the second, for values greater. is undefined, and the graph of has no -intercept.

The -intercept of the graph of a function is the point at which it intersects the -axis. The -coordinate is 0, so the -coordinate can be found by solving the equation

This necessitates setting both definitions of equal to 0 and solving for . For the first definition:

for

Add 16:

By the Square Root Property:

or

Since this definition holds only for , we only select .

For the second definition:

for

Add 25:

By the Square Root Property:

or

Since this definition holds only for , we only select .

Therefore, the graph has two -intercepts, which are at and .

The -intercept of the graph of a function is the point at which it intersects the -axis. The -coordinate is 0, so the -coordinate can be found by solving the equation

This necessitates setting both definitions of equal to 0 and solving for . For the first definition:

for

However, this definition only holds for values less than 0. No solution is yielded.

For the second definition:

for

However, this definition only holds for values greater than or equal to 0. No solution is yielded.

Therefore, has no solution, and the graph of has no -intercepts.

The -intercept of the graph of a function is the point at which it intersects the -axis. The -coordinate is 0, so the -coordinate can be found by evaluating .

The function is piecewise-defined, so it is necessary to use the definition applicable for . Since , and for . this is the definition to use.

The -intercept of the graph is the point .

The general form of the graph of a conic section is

where , , , , and are real coefficients, and and are not both 0.

Here, and .

and are of unlike sign. This indicates that the graph of the equation is a hyperbola.

The general form of the graph of a conic section is

where , , , , and are real coefficients, and and are not both 0.

Here, and .

and and are of like sign. This indicates that the graph of the equation is an ellipse.

The general form of the graph of a conic section is

where , , , , and are real coefficients, and and are not both 0.

Here, since the term is missing, . This indicates that the graph of the equation is a parabola.

The general form of the graph of a conic section is

where , , , , and are real coefficients, and and are not both 0.

Here, and .

and are of unlike sign. This indicates that the graph of the equation is a hyperbola.

If two distinct lines intersect at the point - that is, if both pass through this point - it follows that is a solution of the equations of both. Therefore, set in both equations and determine whether they are true or not.

True - is a solution of this equation.

True - is a solution of this equation.

Therefore, is a solution of both equations, and the lines intersect at this point.

As indicated by the solid line, the graph of the inequality at right includes the line of the equation, so the inequality graphed is either

or

To determine which one, we can select a test point and substitute its coordinates in either inequality, testing whether it is true for those values. The easiest test point is ; it is part of the solution region, so we want the inequality that it makes true. Let us select the first inequality:

makes this inequality true, so the graph of the inequality is the one that includes the origin. This is the correct choice.

(Note that if you select the second inequality, substitution will yield a false statement; this will allow you to draw the same conclusion.)

The graph of an inequality that includes either the or symbol is the graph of the corresponding equation along with all of the points on either side of it. We are given both the line and the circle, so for each inequality, it remains to determine which side of each figure is included. In each case, this can be done by choosing any test point on either side of the figure, substituting its coordinates in the inequality, and determining whether the inequality is true or not. The easiest test point is .

This is true; select the side of this line that includes the origin.

This is true; select the side of this circle that includes the origin - the inside.

The solution sets of the individual inequalities are below:

The graph of the system is the intersection of the two sets, shown below:

As indicated by the dashed line, the graph of the inequality at right does not include the line of the equation, so the inequality graphed is either

or

To determine which one, we can select a test point and substitute its coordinates in either inequality, testing whether it is true for those values. The easiest test point is ; it is not part of the solution region, so we want the inequality that it makes false. Let us select the first inequality:

makes this inequality false, so the graph of the inequality is the one that does not include the origin. This is the correct choice.

(Note that if the second inequality had been selected, would have made it true, so that would not have been the correct choice; we would have again selected the first.)

The circle with center and radius has as its equation, in standard form,

.

is the distance from this center to the point on the circle , which can be calculated using the distance formula

Substitute the coordinates of the points:

We only need to know , which can be set to 25 in the equation. Also, the center being , we can set . The standard form of the equation of the given circle is therefore

.

To find the general form

,

first, expand the squares of the binomials:

Subtract 25 from both sides, and collect like terms

,

the correct general form of the equation.

The circle with center and radius has as its equation, in standard form,

,

so rewrite the equation in this form.

First, rearrange and group the terms as follows:

Complete both perfect square trinomials as follows:

Rewrite the trinomials as the squares of binomials:

Thus, , and the center of the circle is at the point .

We use the distance formula to determine the length of the radius:

Use completing the square to find the standard form of the equation for a circle:

subtract 6 from both sides:

Complete the square for the x and y parts of the equation:

Factor:

From the standard form of the equation of a circle we see that the center is at (3,1) and the radius is 2.