College Algebra - Lines, Circles, and Piecewise Functions

Find the x and y intercepts for

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

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

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

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

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

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

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

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

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

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

Explanation

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

for the x intercept:

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

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).

Which of the following will form a circle when graphed?

$y=\sqrt{3x}-154$

Explanation

Which of the following will form a circle when graphed?

The equation of a circle follows the general formula of

Where r is the radius of the circle, and (h,k) are the coordinates of the center of the circle.

If we examine our answer choices, only

follows this form.

Try without a calculator.

The graph of the equation

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

is which of the following?

Ellipse

Circle

Parabola

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, and .

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

The graph of the equation

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

is which of the following?

Hyperbola

Ellipse

Circle

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, and .

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

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In the above diagram, the line is the graph of the equation

The circle is the graph of the equation

$x^{2}+y^{2}=36$

Graph the system of inequalities

$y \ge 2x-2$

$x^{2}+y^{2} \le 36$

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Explanation

The graph of an inequality that includes either the $\le$ or $\ge$ 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 .

$y \ge 2x-2$

$0 \ge 2 (0)-2$

$0 \ge 0-2$

$0 \ge -2$

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

$x^{2}+y^{2} \le 36$

$0^{2}+0^{2} \le 36$

$0+0 \le 36$

$0\le 36$

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

The solution sets of the individual inequalities are below:

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

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Give the -coordinate(s) of the -intercept(s) of the graph of the function

$f(x) = \left{\begin{matrix} x^{2}- 16 &x < 0 \ x^{2}-25 & x > 0 \end{matrix}\right.$

Explanation

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:

$f(x) = x^{2} - 16$ for

$x^{2} - 16 = 0$

Add 16:

$x^{2} = 16$

By the Square Root Property:

Since this definition holds only for , we only select .

For the second definition:

$f(x) = x^{2} - 25$ for

$x^{2} - 25= 0$

Add 25:

$x^{2} = 25$

By the Square Root Property:

Since this definition holds only for , we only select .

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

The circle on the coordinate plane with center that passes through the point has what equation (general form)?

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

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

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

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

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

Explanation

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

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

is the distance from this center to the point on the circle , 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 , we can set . 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.

The graph of the equation

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

is which of the following?

Parabola

Circle

Hyperbola

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

If (-1,-3) is a point on a circle with its center at (2,5), what is the radius of the circle?

$\sqrt{73}$

$\sqrt{29}$

$\sqrt{57}$

None of these

Cannot be determined

Explanation

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

$D=\sqrt{(x_{1}-x_{2})^{2}+(y_{2}-y_{1})^{2}}$

$D=\sqrt{(-1-2)^2+(-3-5)^2}=\boldsymbol{\sqrt{73}}$

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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?

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

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

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

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

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