SSAT Upper Level Math : How to graph inverse variation

Example Questions

Example Question #1 : How To Graph Inverse Variation

Give the equation of the vertical asymptote of the graph of the equation .

Explanation:

The vertical asymptote of an inverse variation function is the vertical line of the equation , where  is the value for which the expression is not defined. To find , set the denominator to  and solve for :

is the equation of the asymptote.

Example Question #2 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation .

The graph has no -intercept.

The graph has no -intercept.

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set  and solve for :

This is identically false, so the graph has no -intercept.

Example Question #3 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation .

The graph has no -intercept.

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set  and solve for :

is the -intercept.

Example Question #4 : How To Graph Inverse Variation

Give the slope of the line that passes through the - and -intercepts of the graph of the equation

The line cannot exist as described.

The line cannot exist as described.

Explanation:

The graph of  does not have an -intercept. If it did, then it would be the point on the graph with -coordinate 0. If we were to make this substitution, the equation would be

and

This is identically false, so the graph has no -intercept. Therefore, the line cannot exist as described.

Example Question #5 : How To Graph Inverse Variation

Give the -coordinate of a point with a positive -coordinate at which the graphs of the equations  and  intersect.

The graphs of the equations do not intersect.

The graphs of the equations do not intersect.

Explanation:

Substitute  for  in the second equation:

The discriminant of this quadratic expression is , where ; this is

.

The discriminant being negative, there are no real solutions to this quadratic equation. Consequently, there are no points of intersection of the graphs of the two equations on the coordinate plane.

Example Question #6 : How To Graph Inverse Variation

The graphs of the equations  and  intersect in two points; one has a positive -coordinate. and one has a negative -coordinate. Give the -coordinate of the point of intersection that has a positive -coordinate.

Explanation:

Substitute  for  in the second equation:

This quadratic equation can be solved using the  method; the integers with product  and sum 5 are  and 6, so continue as follows:

Either , in which case , or

, in which case

The desired -coordinate is paired with the positive -coordinate, so we substitute 0.5 for  in the first equation:

Example Question #7 : How To Graph Inverse Variation

Give the -coordinate of the point at which the graphs of the equations  and  intersect.

The graphs of the equations do not intersect.

Explanation:

Using the substitution method, set the values of  equal to each other.

Multiply both sides by :

Substitute in either equation:

Example Question #8 : How To Graph Inverse Variation

A line with slope 4 shares its -intercept with that of the graph of the equation . Which of the following is the equation of that line?

This line does not exist, since the graph of  has no -intercept.

Explanation:

The -intercept of the graph of —the point at which it crosses the -axis—is the point at which , so substitute accordingly and solve for :

The -intercept of this graph, and that of the line, is . Since the slope is 4, the slope-intercept form of the equation of the line is

To put it in standard form:

Example Question #9 : How To Graph Inverse Variation

, where  is a right angle, , and . Which of the following cannot be true?

All of the statements given in the other choices are true.

is a right angle

has perimeter

is a right angle

Explanation:

is a right angle and , so

,

making  a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, the length of the short leg  is half that of hypotenuse :

and the length of long leg  is  times that of :

Corresponding sides of congruent triangles are congruent, so ; since , it follows that .

Also, , ,  and , so the perimeter of  is the sum of these, or

.

Corresponding angles are congruent, so  and . By substitution,  and .

The false statement among the choices is that is a right angle.