Coordinate Geometry - GMAT Quantitative

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

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Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Tap to reveal answer

Answer

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

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

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

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Tap to reveal answer

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

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Answer

The -intercepts, if any exist, can be found by setting :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

The only -intercept is .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is .

The correct set of intercepts is $\left \{ (7,0), (0,7) \right \}$ .

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Question

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

Tap to reveal answer

Answer

The -intercepts, if any exist, can be found by setting :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

The only -intercept is .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is .

The correct set of intercepts is $\left \{ (7,0), (0,7) \right \}$ .

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Question

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

Tap to reveal answer

Answer

The -intercepts, if any exist, can be found by setting :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

The only -intercept is .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is .

The correct set of intercepts is $\left \{ (7,0), (0,7) \right \}$ .

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Question

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

Tap to reveal answer

Answer

The -intercepts, if any exist, can be found by setting :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

The only -intercept is .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is .

The correct set of intercepts is $\left \{ (7,0), (0,7) \right \}$ .

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