Graphing - GMAT Quantitative Reasoning

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Question

What is the domain of $y = 4 - x^{2}$ ?

Answer

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

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Question

What is the domain of $y=-2\sqrt{x}$ ?

Answer

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 $x\geq 0$ .

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Question

What is the domain of the function $y=\sqrt{4-x^{2}}$ ?

Answer

To find the domain, we must find the interval on which $\sqrt{4-x^{2}}$ is defined. We know that the expression under the radical must be positive or 0, so $\sqrt{4-x^{2}}$ is defined when $x^{2}\leq 4$ . This occurs when $x \geq -2$ and $x \leq 2$ . In interval notation, the domain is $\left [ -2,2 \right ]$ .

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Question

Define the functions and as follows:

$\small \small f (x) = 2x + \sqrt{x - 1}$

$\small g (x) = 6x - \sqrt{x-1}$

What is the domain of the function ?

Answer

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,

$\small \small x - 1\geq 0$ , or

$\small \small \small x \geq 1$

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is $\small \small [1, \infty )$

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Question

Define $f (x) = \frac{1}{x ^{2}- 25 }$ .

What is the natural domain of ?

Answer

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:

$x ^{2}-25 =0$

$x ^{2}= 25$

$x = -5 \textrm{ or }x = 5$

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

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$ .

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Question

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

Answer

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

$\sqrt[3]{x -27} = 0$

$\left (\sqrt[3]{x -27} \right )^{3}= 0^{3}$

27 is the only number excluded from the domain.

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Question

Define $g(x) = \frac{x}{x^{2}+3x-4 }$

What is the natural domain of ?

Answer

Since the expression $x^{2}+3x-4$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which $x^{2}+3x-4 =0$ . We solve for by factoring the polynomial, which we can do as follows:

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

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

$x + 4 = 0 \Rightarrow x = -4$

$x -1 = 0 \Rightarrow x = 1$

Therefore, the domain excludes these two values of .

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Question

A line has slope 4. Which of the following could be its - and -intercepts, respectively?

Answer

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

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

and

Slope: $- \left (b\div a \right )=- \left (3.2\div 12.8 \right )=-0.25$

and

Slope: $- \left (b\div a \right )=- \left (17.2\div 4.3 \right )=-4$

and

Slope: $- \left (b\div a \right )= - \left (-3.1 \div 12.4 \right )= 0.25$

and

Slope: $- \left (b\div a \right )=- \left (-10.8\div 2.7 \right )=4$

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

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Question

A line has slope $\frac{3}{4}$ . Which of the following could be its - and -intercepts, respectively?

Answer

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

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 $- (b\div a)$ is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope $\frac{3}{4}$ .

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Question

A line has slope $-\frac{4}{3}$ . Which of the following could be its - and -intercepts, respectively?

Answer

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

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

$\left ( \frac{5}{6}, 0 \right )$ and $\left ( 0, \frac{5}{8}\right )$ :

Slope: $- \left (b\div a \right )=- \left (\frac{5}{8}\div \frac{5}{6} \right )= - \frac{3}{4}$

$\left ( \frac{6}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

Slope: $- \left (b\div a \right )=- \left ( \frac{4}{7}\div \frac{6}{7} \right )=-\frac{2}{3}$

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

Slope: $- \left (b\div a \right )= - \left ( \frac{4}{7}\div \frac{3}{7} \right )= -\frac{4}{3}$

$\left ( \frac{4}{5}, 0 \right )$ and $\left ( 0, \frac{3}{5}\right )$

Slope: $- \left (b\div a \right )=- \left ( \frac{3}{5}\div \frac{4}{5} \right )= -\frac{3}{4}$

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$ comprise the correct choice.

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Question

Which of the following equations can be graphed with a line perpendicular to the green line in the above figure, and with the same -intercept?

Answer

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 $-\frac{b}{a}$ . This makes the slope of the green line $-\frac{5}{3}$ .

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or $\frac{3}{5}$ . Since the desired line must also have -intercept , then the slope-intercept form of the line is

$y = \frac{3}{5}x+ 5$

which can be rewritten in standard form:

$5y =5 \left ( \frac{3}{5}x+ 5 \right )$

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Question

Which of the following equations can be graphed with a line perpendicular to the green line in the above figure, and with the same -intercept?

Answer

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 $-\frac{b}{a}$ . This makes the slope of the green line $-\frac{5}{3}$ .

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or $\frac{3}{5}$ . Since the desired line must also have -intercept , the equation of the line, in point=slope form, is

$y - y_{1} = m (x-x_{1})$

$y -0 = \frac{3}{5} (x-3)$

which can be simplified as

$y = \frac{3}{5} x-\frac{9}{5}$

$5 \cdot y =5 \cdot \left ( \frac{3}{5} x-\frac{9}{5} \right )$

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Question

Which of the following equations can be graphed with a line parallel to the green line in the above figure?

Answer

If and be the - and -intercepts, respectively, of a line, the slope of the line is $-\frac{b}{a}$ .

The - and -intercepts of the line are, respectively, and , so , and consequently, the slope of the green line is $-\frac{5}{3}$ . A line parallel to this line must also have slope $m = -\frac{5}{3}$ .

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 $-\frac{5}{3}$ as its -coefficient is $y = -\frac{5}{3}x - 7$ , so this is the correct choice.

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Question

The graph of the equation $y = x^{2}+7x - 18$ shares its -intercept and one of its -intercepts with a line of positive slope. What is the equation of the line?

Answer

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:

$y = x^{2}+7x - 18$

$y = 0^{2}+7 \cdot 0 - 18$

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:

$y = x^{2}+7x - 18$

$x^{2}+7x - 18 = 0$

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 $-\frac{b}{a}$ . If the intercepts are and , the slope is $-\frac{-18}{2}= 9$ ; if the intercepts are and , the slope is $-\frac{-18}{-9}= -2$ . 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, .

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Question

The graph of the equation $y =-6x^{2}+ 11x + 10$ shares its -intercept and one of its -intercepts with a line of negative slope. Give the equation of that line.

Answer

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:

$y =-6x^{2}+ 11x + 10$

$y =-6 \cdot 0^{2}+ 11 \cdot 0+ 10$

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:

$y =-6x^{2}+ 11x + 10$

$-6x^{2}+ 11x + 10= 0$

$-\left (-6x^{2}+ 11x + 10 \right )=- 0$

$6x^{2}- 11x -10= 0$

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:

$6x^{2}-15x+4 x -10= 0$

$(6x^{2}-15x)+(4 x -10)= 0$

Split:

$x= -\frac{2}{3}$

or

$x = \frac{5}{2}$

The -intercepts of the parabola are $\left ( -\frac{2}{3}, 0 \right )$ and $\left ( \frac{5}{2}, 0 \right )$ , 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 $-\frac{b}{a}$ , or, equivalently, $- \left ( b \div a\right )$

If the intercepts are $\left ( -\frac{2}{3}, 0 \right )$ and , the slope is $- \left [ 10 \div \left (- \frac{2}{3} \right )\right ] = 15$ ; if the intercepts are $\left ( \frac{5}{2}, 0 \right )$ and , the slope is $- \left ( 10 \div \frac{5}{2}\right ) = -4$ . 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, .

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Question

The graph of the equation $x = y ^{2} - y - 12$ shares its -intercept and one of its -intercepts with a line of positive slope. What is the equation of the line?

Answer

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:

$x = y ^{2} - y - 12$

$x =0 ^{2} - 0- 12$

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:

$x = y ^{2} - y - 12$

$y ^{2} - y - 12 = 0$

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 $-\frac{b}{a}$ . If the intercepts of the line are and , the slope of the line is $-\frac{-3}{-12} = -\frac{1}{4}$ ; if the intercepts are and , the slope is $-\frac{4}{-12} = \frac{1}{3}$ . We choose the latter, since we are looking for a line with positive slope; since its -intercept is , then we can substitute $m = \frac{1}{3}, b = 4$ in the slope-intercept form of the line, , to get the correct equation, $y = \frac{1}{3}x+4$ .

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Question

Give the -coordinate of the point of intersection of the lines of the equations:

Round your answer to the nearest whole number, if applicable.

Answer

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:

$\underline{16x+28y = 128}$

$x = 576 \div 65 \approx 8.9$ ,

making 9 the correct response.

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Question

Which of these equations is represented by a line that does not intersect the graph of the equation $y = 2x^{2} - 7x+5$ ?

Answer

We can find out whether the graphs of and $y = 2x^{2} - 7x+5$ intersect by first solving for in the first equation:

We then substitute in the second equation for :

$2x^{2} - 7x+5 = 2-x$

Then we rewrite in standard form:

$2x^{2} - 7x+5 -2 + x = 2-x -2 + x$

$2x^{2} - 6x+3 = 0$

Since we are only trying to pdetermine whether at least one point of intersection exists, rather than actually find the point, all we need to do is to evaluate the discriminant; if it is nonnegative, at least one solution - and, consequently, one point of intersection - exists. In the general quadratic equation $ax^{2} +bx+c = 0$ , this is $b^{2} - 4ac$ , so here, the discriminant is

$b^{2} - 4ac = (-6)^{2}- 4(2)(3) = 36-24 = 12\ge 0$ .

Therefore, the line of the equation intersects the parabola of the equation $y = 2x^{2} - 7x+5$ .

We do the same for the other three lines:

$2x^{2} - 7x+5 = 4-x$

Then we rewrite in standard form:

$2x^{2} - 7x+5 -4+ x = 4-x -4 + x$

$2x^{2} - 6x+1 = 0$

$b^{2} - 4ac = (-6)^{2}- 4(2)(1) = 36-8= 28\ge 0$ .

The line of intersects the parabola.

$2x^{2} - 7x+5 = x-2$

$2x^{2} - 7x+5- x+2 = x-2 - x+2$

$2x^{2} - 8x+7 =0$

$b^{2} - 4ac = (-8)^{2}- 4(2)(7) = 64-56= 8\ge 0$

The line of intersects the parabola.

$2x^{2} - 7x+5 = x-4$

$2x^{2} - 7x+5- x+4 = x-4 - x+4$

$2x^{2} - 8x+9 =0$

$b^{2} - 4ac = (-8)^{2}- 4(2)(9) = 64-72= -8 < 0$

Since the discriminant is negative, the system has no real solution. This means that the line of does not intersect the parabola of the equation $y = 2x^{2} - 7x+5$ , and it is the correct choice.

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Question

A line passes through the vertex and the -intercept of the parabola of the equation $y = x^{2} - 8x+ 15$ . What is the equation of the line?

Answer

To locate the -intercept of the equation $y = x^{2} - 8x+ 15$ , substitute 0 for :

$y = x^{2} - 8x+ 15$

$y = 0^{2} - 8 \cdot 0+ 15$

The -intercept of the parabola is .

The vertex of the parabola of an equation of the form $y = ax^{2}+ bx + c$ has -coordinate $-\frac{b}{2a}$ . Here, we substitute , to obtain -coordinate

$-\frac{b}{2a} = -\frac{-8}{2 (1)} = 4$ .

To find the -coordinate, substitute this for :

$y = x^{2} - 8x+ 15$

$y = 4^{2} - 8 \cdot 4+ 15 = 16 - 32 +15 = -1$

The vertex is .

The line includes points and ; apply the slope formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m = \frac{15-(-1)}{0-4} = \frac{16}{-4} = -4$

The slope is , and the -intercept is ; in the slope-intercept form , substitute for and . The equation of the line is .

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Question

A line with positive slope passes through the vertex and an -intercept of the parabola of the equation $y = x^{2} + 12x + 20$ . What is the equation of the line?

Answer

The vertex of the parabola of an equation of the form $y = ax^{2}+ bx + c$ has -coordinate $-\frac{b}{2a}$ . Here, we substitute , to obtain -coordinate

$-\frac{b}{2a} = -\frac{12}{2 (1)} = -6$ .

To find the -coordinate, substitute this for :

$y = x^{2} + 12x + 20$

$y = (-6)^{2} + 12(-6) + 20 = 36 -72 + 20 = -16$

The vertex is .

To find the -intercepts of the parabola, substitute 0 for in the equation:

$x^{2} + 12x + 20 = 0$

Either

, in which case ,

or

, in which case .

The -intercepts are and .

The line includes and either or , so we find the slope in each case using the slope formula.

If the line includes and :

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{0 - (-16)}{-2 - (-6)} = \frac{16}{4} = 4$ .

If the line includes and :

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{0 - (-16)}{-10 - (-6)} = \frac{16}{-4} = -4$

We choose the first case, since the line has positive slope. The line through and has as its equation, using the point-slope form with and point :

$y-y_{1} = m (x-x_{1})$

$y-0= 4\left [ (x-(-2) \right ]$

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