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College Algebra : Lines, Circles, and Piecewise Functions

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Example Questions

Example Question #91 : College Algebra

Untitled

Refer to the figures above.

At left is the graph of the equation $y = -\frac{1}{2}x+ 5$ . Which inequality is graphed at right?

Possible Answers:

$y < -\frac{1}{2}x+ 5$

$y \ge -\frac{1}{2}x+ 5$

$y \le -\frac{1}{2}x+ 5$

$y > -\frac{1}{2}x+ 5$

Correct answer:

$y \le -\frac{1}{2}x+ 5$

Explanation:

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

$y \le -\frac{1}{2}x+ 5$

$y \ge -\frac{1}{2}x+ 5$

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 (0,0) ; it is part of the solution region, so we want the inequality that it makes true. Let us select the first inequality:

$y \le -\frac{1}{2}x+ 5$

$0 \le -\frac{1}{2} (0)+ 5$

$0 \le0 + 5$

$0 \le 5$

(0,0) makes this inequality true, so the graph of the inequality $y \le -\frac{1}{2}x+ 5$ 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.)

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Example Question #92 : College Algebra

The graph of the equation

$3y^{2} +12x-4y-42=0$

is which of the following?

Possible Answers:

Ellipse

Hyperbola

Circle

Parabola

Correct answer:

Parabola

Explanation:

The general form of the graph of a conic section is

$Ax^{2}+ By^{2}+ Cx + Dy + E= 0$

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

Here, since the $x^{2}$ term is missing, A= 0 . This indicates that the graph of the equation is a parabola.

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Example Question #12 : Lines, Circles, And Piecewise Functions

The graph of the equation

$x^{2}-y^{2}-8x+16y+34=0$

is which of the following?

Possible Answers:

Hyperbola

Ellipse

Parabola

Circle

Correct answer:

Hyperbola

Explanation:

The general form of the graph of a conic section is

$Ax^{2}+ By^{2}+ Cx + Dy + E= 0$

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

Here, A = 1 and B = -1 .

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

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Example Question #93 : College Algebra

The graph of the equation

$3x^{2}-4y^{2}+12x-4y-42=0$

is which of the following?

Possible Answers:

Ellipse

Hyperbola

Circle

Parabola

Correct answer:

Hyperbola

Explanation:

The general form of the graph of a conic section is

$Ax^{2}+ By^{2}+ Cx + Dy + E= 0$

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

Here, A = 3 and B = -4 .

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

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Example Question #11 : Graphs

Untitled

Refer to the figures above.

At left is the graph of the equation $y = \frac{4}{3} x- 8$ . Which inequality is graphed at right?

Possible Answers:

$y < \frac{4}{3} x- 8$

$y \ge \frac{4}{3} x- 8$

$y \le \frac{4}{3} x- 8$

$y > \frac{4}{3} x- 8$

Correct answer:

$y < \frac{4}{3} x- 8$

Explanation:

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

$y < \frac{4}{3} x- 8$

$y > \frac{4}{3} x- 8$

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 (0,0) ; it is not part of the solution region, so we want the inequality that it makes false. Let us select the first inequality:

$y < \frac{4}{3} x- 8$

$0 < \frac{4}{3} (0)- 8$

0 < 0- 8

0 < -8

(0,0) makes this inequality false, so the graph of the inequality $y < \frac{4}{3} x- 8$ is the one that does not include the origin. This is the correct choice.

(Note that if the second inequality had been selected, (0,0) would have made it true, so that would not have been the correct choice; we would have again selected the first.)

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Example Question #11 : Lines, Circles, And Piecewise Functions

Try without a calculator.

The graph of the equation

$3x^{2}+4y^{2}+12x-4y-42=0$

is which of the following?

Possible Answers:

Parabola

Circle

Hyperbola

Ellipse

Correct answer:

Ellipse

Explanation:

The general form of the graph of a conic section is

$Ax^{2}+ By^{2}+ Cx + Dy + E= 0$

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

Here, A = 3 and B = 4 .

$A \ne B$ and and are of like sign. This indicates that the graph of the equation is an ellipse.

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Example Question #92 : College Algebra

The circle on the coordinate plane with center (6, -7 ) that passes through the point (2, -4 ) has what equation (general form)?

Possible Answers:

$x^{2}+ y^{2} - 14x + 12y+60 = 0$

$x^{2}+ y^{2} - 14x - 12y+60 = 0$

$x^{2}+ y^{2} - 12 x + 14 y+60 = 0$

$x^{2}+ y^{2} +14x - 12y+60 = 0$

$x^{2}+ y^{2} + 12 x - 14 y+60 = 0$

Correct answer:

$x^{2}+ y^{2} - 12 x + 14 y+60 = 0$

Explanation:

The circle with center (h,k ) and radius has as its equation, in standard form,

$(x- h)^{2}+ (y-k)^{2} = r^{2}$ .

is the distance from this center (6, -7 ) to the point on the circle (2, -4 ) , which can be calculated using the distance formula

$r^{2} = (x_{1} - x_{2}) ^{2} + (y_{1} - y_{2}) ^{2}$

Substitute the coordinates of the points:

$r^{2} = (6-2) ^{2} + (-7-(-4)) ^{2}$

$r^{2} = 4^{2} + (-3) ^{2}$

$r^{2} = 16+9$

$r^{2} = 25$

We only need to know $r^{2}$ , which can be set to 25 in the equation. Also, the center being (6, -7 ) , we can set h = 6, k = -7 . The standard form of the equation of the given circle is therefore

$(x- 6)^{2}+ (y+7)^{2} =25$ .

To find the general form

$x^{2}+ y^{2}+ Cx + Dy + E = 0$ ,

first, expand the squares of the binomials:

$x^{2}- 2 \cdot x \cdot 6 + 6^{2}+ y^{2}+ 2 \cdot y \cdot 7 + 7^{2} = 25$

$x^{2}- 12 x +3 6+ y^{2}+ 14y + 49 = 25$

Subtract 25 from both sides, and collect like terms

$x^{2}- 12 x +3 6+ y^{2}+ 14y + 49 - 25 = 25 - 25$

$x^{2}- 12 x + y^{2}+ 14 y+60 = 0$

$x^{2}+ y^{2} - 12 x + 14 y+60 = 0$ ,

the correct general form of the equation.

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Example Question #12 : Graphs

Give the coordinates of the center of the circle of the equation

$x^{2}+ y^{2}+ 12x + 16y + 30 = 0$

Possible Answers:

(0,0)

(6, 8)

(12, 16)

(-12, -16)

(-6, -8)

Correct answer:

(-6, -8)

Explanation:

The circle with center (h,k ) and radius has as its equation, in standard form,

$(x- h)^{2}+ (y-k)^{2} = r^{2}$ ,

so rewrite the equation in this form.

First, rearrange and group the terms as follows:

$x^{2}+ y^{2}+ 12x + 16y + 30 = 0$

$x^{2}+ 12x + y^{2}+ 16y + 30 = 0$

$(x^{2}+ 12x )+( y^{2}+ 16y )= -30$

Complete both perfect square trinomials as follows:

$(x^{2}+ 12x+ 36 )+( y^{2}+ 16y+ 64)= -30 + 36 + 64$

$(x^{2}+ 12x+ 36 )+( y^{2}+ 16y+ 64)=70$

Rewrite the trinomials as the squares of binomials:

$(x+6)^{2}+ (y+8)^{2}= 70$

Thus, h = -6, k= -8 , and the center of the circle is at the point (-6, -8) .

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Example Question #11 : Graphs

What is the center and radius of a circle whose equation is given by: x^2-6x+y^2-2y+6=0

Possible Answers:

r=5

(h,k)=(5,5)

r=3

(h,k)=(1,1)

r=1

(h,k)=(1,3)

r=2

(h,k)=(3,1)

r=5

(h,k)=(1,5)

Correct answer:

r=2

(h,k)=(3,1)

Explanation:

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

x^2-6x+y^2-2y+6=0

subtract 6 from both sides:

x^2-6x+y^2-2y=-6

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

x^2-6x+9+y^2-2y+1=-6+10

Factor:

(x-3)^2+(y-1)^2=4

(x-h)^2+(y-k)^2=r^2

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

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Example Question #101 : College Algebra

Find the x and y intercepts for (x-6)^2+(y+3)^2=16

Possible Answers:

$x \hspace{1mm}intercept= none$

$y\hspace{1mm}intercept= none$

$x \hspace{1mm}intercept= -6\pm \sqrt{7}$

$y\hspace{1mm}intercept= none$

$x \hspace{1mm}intercept= 6\pm \sqrt{6}$

$y\hspace{1mm}intercept= none$

$x \hspace{1mm}intercept= none$

$y\hspace{1mm}intercept= 6\pm\sqrt{5}$

$x \hspace{1mm}intercept= 6\pm \sqrt{7}$

$y\hspace{1mm}intercept= none$

Correct answer:

$x \hspace{1mm}intercept= 6\pm \sqrt{7}$

$y\hspace{1mm}intercept= none$

Explanation:

To find the x or y intercept in any equation we set y=0 or x=0 respectively.

for the x intercept:

(x-6)^2+(3)^2=16

(x-6)^2=7

$\boldsymbol{x=6\pm \sqrt{7}}$

The y intercept does not exist because to solve this equation for y requires complex roots (there will be a negative sign under the radical).

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College Algebra : Lines, Circles, and Piecewise Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #91 : College Algebra

Example Question #92 : College Algebra

Example Question #12 : Lines, Circles, And Piecewise Functions

Example Question #93 : College Algebra

Example Question #11 : Graphs

Example Question #11 : Lines, Circles, And Piecewise Functions

Example Question #92 : College Algebra

Example Question #12 : Graphs

Example Question #11 : Graphs

Example Question #101 : College Algebra

All College Algebra Resources

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