Quadratic Inequalities - Algebra II

Rewrite in standard form and factor:

The zeroes of the polynomial are therefore , so we test one value in each of three intervals , , and to determine which ones are included in the solution set.

:

Test :

False; is not in the solution set.

:

Test

True; is in the solution set

:

Test :

False; is not in the solution set.

Since the inequality symbol is , the boundary points are not included. The solution set is the interval .

The first step of questions like this is to get the quadratic in its standard form. So we move the over to the left side of the inequality:

This quadratic can easily be factored as. So now we can write this in the form

and look at each of the factors individually. Recall that a negative number times a negative is a positive number. Therefore the boundaries of our solution interval is going to be when both of these factors are negative. is negative whenever , and is negative whenever . Since , one of our boundaries will be . Remember that this will be an open interval since it is less than, not less than or equal to.

Our other boundary will be the other point when the product of the factors becomes positive. Remember that is positive when , so our other boundary is . So the solution interval we arrive at is

When asked to solve for x we need to isolate x on one side of the equation.

To do this our first step is to subtract 7 from both sides.

From here, we divide by 4 to solve for x.

When asked to solve for y we need to isolate the variable on one side and the constants on the other side.

To do this we first add 9 to both sides.

From here, we divide by -12 to solve for y.

The region contains only values which are greater than or equal to those on the line , so its values are .

Similarly, the region contains only values which are less than or equal to those on the line , so its values are .

The region contains only values which are greater than or equal to those on the line , so its values are .

Also, the region contains only values which are less than or equal to those on the line , so its values are .

To begin, we analyze the equation given: the base equation, is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows:

Simplified as:

Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph:

The hyperbola in the question's graph is .

This could be discerned because it subtracts from and not the other way around, putting the zeros on the axis. This narrows the answers down to and .

Test a point to tell which inequality is being used in the graph.

The point is within the shaded region, for example.

, which is greater than 16, so the answer is .

The hyperbola in the graph has y-intercepts rather than x-intercepts, so the equation must be in the form and not the other way around.

The y-intercepts are at 1 and -1, so the correct equation will have just and not .

The answer not must either be,

or .

To see which, test a point in the shaded area.

For example, .

, which is less than 1, so the answer is .

The equation for a horizontal hyperbola is. The equation for a vertical hyperbola is . Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The fact that the right side of the inequality is not equal to 1 does not change the fact that , and represent hyperbolas, since THESE can all be simplified to create an inequality with 1 on the right side (by dividing both sides of the equation by the constant on the right side of the inequality.) Answer choice is the only option in which the two terms on the left side of the inequality are combined using addition rather than subtraction, creating an ellipse rather than a hyperbola. (The equation for an ellipse is .)

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The presence of coefficients in and does not change the fact that and represent hyperbolas, since both can be simplified to remove those coefficients (by dividing the numerator and denominator of terms with coefficients by those coefficients.) Answer choice is missing an exponent of 2 on the first term in the inequality, and therefore does not match the form of a hyperbola.

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . In both, is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The graph shows a vertical hyperbola, so in its corresponding inequality the y-term must appear first. The center is shaded, so the left side of the graph’s corresponding inequality (the side containing the variables x and y) is less than the constant on the right side. The lines are dashed rather than solid, so the inequality sign must be rather than .

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The graph shows a horizontal hyperbola, so in its corresponding inequality the x-term must appear first. The center is not shaded, so the left side of the graph’s corresponding inequality (the side containing the variables x and y) is greater than the constant on the right side. The lines are dashed rather than solid, so the inequality sign must be rather than . The center lies at (-1, 1), so x must be followed by the constant 1, and y must be followed by the constant -1.

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The graph shows a horizontal hyperbola, so in its corresponding inequality the x-term must appear first. The center is not shaded, so the left side of the graph’s corresponding inequality (the side containing the variables x and y) is greater than the constant on the right side.

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The graph shows a vertical hyperbola, so in its corresponding inequality the y-term must appear first. The center is (0, 0), so neither x nor y can be followed by a constant. The center is shaded, so the left side of the graph’s corresponding inequality (the side containing the variables x and y) is less than the constant on the right side. The lines are solid rather than dashed, so the inequality sign must be rather than .

The equation for a horizontal hyperbola is. The equation for a vertical hyperbola is . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. Neither the x nor the y in the inequality are followed by constants, so the graph must be centered on (0,0). The x-term appears first, so the graph must show a horizontal hyperbola. The inequality sign is rather than , so the lines must be dashed rather than solid. The left side is less than rather than greater than the constant, so the center must be shaded.

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The x and y in the inequality are followed by the constants 2 and -1 respectively, so the graph must be centered on (-2, 1). The y-term appears first, so the graph must show a vertical hyperbola. The inequality sign is rather than , so the lines must be solid rather than dashed. The left side is greater than rather than less than the constant, so the center must not be shaded.

The hyperbola in the question's graph is .

This could be discerned because it subtracts from and not the other way around, putting the zeros on the axis. This narrows the answers down to and .

Test a point to tell which inequality is being used in the graph.

The point is within the shaded region, for example.

, which is greater than 16, so the answer is .

The hyperbola in the graph has y-intercepts rather than x-intercepts, so the equation must be in the form and not the other way around.

The y-intercepts are at 1 and -1, so the correct equation will have just and not .

The answer not must either be,

or .

To see which, test a point in the shaded area.

For example, .

, which is less than 1, so the answer is .

The equation for a horizontal hyperbola is. The equation for a vertical hyperbola is . Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. The fact that the right side of the inequality is not equal to 1 does not change the fact that , and represent hyperbolas, since THESE can all be simplified to create an inequality with 1 on the right side (by dividing both sides of the equation by the constant on the right side of the inequality.) Answer choice is the only option in which the two terms on the left side of the inequality are combined using addition rather than subtraction, creating an ellipse rather than a hyperbola. (The equation for an ellipse is .)