Graphing - GMAT Quantitative

Set in the equation and solve for .

The -intercept is

To find the horizontal asymptote, we can divide both numerator and denominator in the right expression by :

As approaches positive or negative infinity, and both approach 0. Therefore, approaches , making the horizontal asymptote the line of the equation .

The domain of the function specifies the values that can take. Here, is defined for every value of , so the domain is all real numbers.

To find the domain, we need to decide which values can take. The is under a square root sign, so cannot be negative. can, however, be 0, because we can take the square root of zero. Therefore the domain is .

To find the domain, we must find the interval on which is defined. We know that the expression under the radical must be positive or 0, so is defined when . This occurs when and . In interval notation, the domain is .

The domain of is the intersection of the domains of and . and are each restricted to all values of that allow the radicand to be nonnegative - that is,

, or

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is

The only restriction on the domain of is that the denominator cannot be 0. We set the denominator to 0 and solve for to find the excluded values:

The domain is the set of all real numbers except those two - that is,

.

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

27 is the only number excluded from the domain.

Since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which . We solve for by factoring the polynomial, which we can do as follows:

Replacing the question marks with integers whose product is and whose sum is 3:

Therefore, the domain excludes these two values of .

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We do not need to find the actual slopes of the four choices if we observe that in each case, and are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope .

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice, since a line passing through these points has the correct slope.

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and :

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice.

The - and -intercepts of the line are, respectively, and . If and are the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line .

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or . Since the desired line must also have -intercept , then the slope-intercept form of the line is

which can be rewritten in standard form:

The slope of the green line can be calculated by noting that the - and -intercepts of the line are, respectively, and . If and be the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line .

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or . Since the desired line must also have -intercept , the equation of the line, in point=slope form, is

which can be simplified as

If and be the - and -intercepts, respectively, of a line, the slope of the line is .

The - and -intercepts of the line are, respectively, and , so , and consequently, the slope of the green line is . A line parallel to this line must also have slope .

Each of the equations of the lines is in slope-intercept form , where is the slope, so we need only look at the coefficients of . The only choice that has as its -coefficient is , so this is the correct choice.

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

The quadratic expression can be "reverse-FOILed" by noting that 9 and have product and sum 7:

, in which case

or

, in which case .

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We will examine both possibilities

If and be the - and -intercepts, respectively, of the line, then the slope of the line is . If the intercepts are and , the slope is ; if the intercepts are and , the slope is . Since the line is of positive slope, we choose the line of slope 9; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

Using the method, split the middle term by finding two integers whose product is and whose sum is ; by trial and error we find these to be and 4, so proceed as follows:

Split:

or

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We examine both possibilities.

If and be the - and -intercepts, respectively, of the line, then the slope of the line is , or, equivalently,

If the intercepts are and , the slope is ; if the intercepts are and , the slope is . Since the line is of negative slope, we choose the line of slope ; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a horizontal parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

Either

, in which case ,

or

, in which case .

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We will examine both possibilities.

If and be the - and -intercepts, respectively, of the line, then the slope of the line is . If the intercepts of the line are and , the slope of the line is ; if the intercepts are and , the slope is . We choose the latter, since we are looking for a line with positive slope; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The point of intersection of the two lines has as its coordinates the values of and that make both of the given linear equations true. Therefore, we seek to find the solution of the system of equations:

We need only find , so multiply both sides of the two equations by 7 and 4, respectively. Then add:

,

making 9 the correct response.