DSQ: Graphing a quadratic function - GMAT Quantitative

The two statements together give the vertex of the parabola, but no clue is given as to the orientation of the parabola - horizontal, vertical, or otherwise. Without that information, or any way to find it, the line of symmetry cannot be determined.

The line of symmetry of a horizontal parabola with vertex at is the horizonal line of the equation . In other words, the -coordinate of the vertex, which is given in Statement 2 but not Statement 1, is the one and only thing needed.

Assume both statements are true, and consider these two equations:

Case 1:

The -intercept can be proven to be by substituting 0 for :

We show that the graph intersects the line of equation twice by substituting 3 for :

The points of intersection are .

To find the -intercept(s), if any exist, substitute 0 for :

This has no real solutions, so the parabola has no -intercepts.

Case 2:

The -intercept be proved to be by substituting 0 for :

We show that the graph intersects the line of equation twice by substituting 3 for :

We examine the discriminant:

The discriminant is positive, so there are two real solutions, meaning that there are two points of intersection.

To find the -intercept(s), if any exist, substitute 0 for :

We examine this discriminant:

The discriminant is positive, so there are two real solutions, meaning that there are two -intercepts.

Two parabolas have been identified fitting the main condition and those of both statements, but one has no -intercept and one has two. The two statements together provide insufficient information.

From Statement 1, since the vertex is not on the -axis - its -coordinate is nonzero - the parabola has either zero or two -intercepts. However, with no further information, it is not possible to choose. Statement 2 alone is not helpful since it only gives one point, and no further information about it.

Assume both statements to be true. We can find the equation of the parabola as follows:

A parabola with vertex has equation

for some nonzero .

From Statement 1, , so the equation becomes

Since the parabola passes through To find , we substitute 0 for and 21 for :

The equation of the parabola is .

Now that the equation is known, the -intercept(s) themselves, if any, can be found by substituting 0 for .

Assume Statement 1 alone. By vertical symmetry, if two points of a parabola have the same -coordinate, the line of symmetry is the vertical line that passes halfway between them. and have the same -coordinate, so the axis of symmetry must be

, or .

Statement 1 alone is sufficient.

Statement 2 can be proved sufficient using a similar argument.

The line of symmetry of a vertical parabola is the vertical line passing through the vertex. Each statement alone gives only one point on the graph, neither of which is the vertex, so neither statement alone gives sufficient information.

Now assume both statements to be true. Statement 1 gives one -intercept, ; Statement 2 states that the graph passes through the origin, so it is not only the -intercept, it is also the other -intercept. The -coordinate of the vertex is the arithmetic mean of those of the two -intercepts, so that value is

Only the -coordinate of the vertex is needed to answer the question - we can immediately deduce that the line of symmetry is .

The line of symmetry of a vertical parabola is the vertical line passing through the vertex. Statement 1 alone is not helpful, since it only gives the -intercept.

Statement 2 alone, however, answers the question. In a parabola with only one -intercept, that -intercept, given in Statement 2 as , doubles as the vertex. The vertical line through the vertex, which here is the line with equation , is the line of symmetry.

Assume Statement 1 alone. The number of -intercepts of the graph of the function - depends on the sign of the discriminant of the expression, .

If , then the discriminant becomes

Since in a quadratic equation, is nonzero, must be positive, and discriminant must be negative. This means that the parabola of has no -intercepts.

We show that Statement 2 alone gives insufficient information by examining two equations: and . In both equations, the sum of the coefficients is 9.

In the first equation, the discriminant is

, a positive value, so the parabola of has two -intercepts.

In the second equation, however, the discriminant is

, a negative value, so the parabola of has no -intercepts.

The parabola is concave upward if and only if , and concave downward if and only if . Therefore, we need to know the sign of to answer the question. We show that the two statements together provide insufficient information by examining two equations.

Case 1:

; we check the values of the expressions in both statements:

The conditions of both statements are satisfied; since , the parabola that graphs this function is concave upward.

Case 2:

; we check the values of the expressions in both statements:

The conditions of both statements are satisfied; since , the parabola that graphs this function is concave downward.

Assume Statement 1 alone. The number of -intercepts(s) of the graph of depends on the sign of discriminant . By Statement 1, , or, equivalently, , which means that the parabola of has exactly one -intercept.

Statement 2 alone, that the quadratic coefficient is positive, only establishes that the parabola is concave upward. Therefore, it gives insufficient information.

The parabola is concave upward if and only if , and concave downward if and only if . Therefore, we need to know the sign of to answer the question. Statement 2, but not Statement 1, gives us the value of , the sign of which is positive, so Statement 2 alone, but not Statement 1 alone, tells us the parabola is concave upward.

The line of symmetry of a vertical parabola with vertex at has as its equation . In other words, the -coordinate, which is given in Statement 1 but not Statement 2, is the one and only thing needed.

Assume Statement 1 alone. Since - that is, the function has a negative discriminant - the graph of has no -intercepts. This alone, however, does not determine whether the parabola is concave upward or concave downward. Also, Statement 2 alone only gives one point of the parabola, thereby providing insufficient information.

Now assume both statements are true. From Statement 2, , so the parabola has a point above the -axis. If the parabola is concave downward, then it must cross the -axis, which is impossible as a result of Statement 1. The parabola therefore must be concave upward.

Assume both statements are true.

The parabola is concave upward if and only if , and concave downward if and only if . Therefore, we need to know the sign of to answer the question. We show that the two statements together provide insufficient information by examining two equations.

Case 1:

If we substitute 0 for , is easily seen to be equal to 6, so the -intercept is . Also:

Case 2:

If we substitute 0 for , is easily seen to be equal to 6, so the -intercept is . Also:

Each equation satisfies the conditions of both statements, but is positive in one equation and negative in the other. can assume either sign, so the question of the parabola's concavity is unresolved.

Whether the parabola of a quadratic function is concave upward or concave downward depends on one thing and one thing only - whether quadratic coefficient is positive or negative. Statement 1 gives you this information; Statement 2 does not.

The two statements together give the vertex of the parabola, but no clue is given as to the orientation of the parabola - horizontal, vertical, or otherwise. Without that information, or any way to find it, the line of symmetry cannot be determined.

The line of symmetry of a horizontal parabola with vertex at is the horizonal line of the equation . In other words, the -coordinate of the vertex, which is given in Statement 2 but not Statement 1, is the one and only thing needed.

Assume both statements are true, and consider these two equations:

Case 1:

The -intercept can be proven to be by substituting 0 for :

We show that the graph intersects the line of equation twice by substituting 3 for :

The points of intersection are .

To find the -intercept(s), if any exist, substitute 0 for :

This has no real solutions, so the parabola has no -intercepts.

Case 2:

The -intercept be proved to be by substituting 0 for :

We show that the graph intersects the line of equation twice by substituting 3 for :

We examine the discriminant:

The discriminant is positive, so there are two real solutions, meaning that there are two points of intersection.

To find the -intercept(s), if any exist, substitute 0 for :

We examine this discriminant:

The discriminant is positive, so there are two real solutions, meaning that there are two -intercepts.

Two parabolas have been identified fitting the main condition and those of both statements, but one has no -intercept and one has two. The two statements together provide insufficient information.

From Statement 1, since the vertex is not on the -axis - its -coordinate is nonzero - the parabola has either zero or two -intercepts. However, with no further information, it is not possible to choose. Statement 2 alone is not helpful since it only gives one point, and no further information about it.

Assume both statements to be true. We can find the equation of the parabola as follows:

A parabola with vertex has equation

for some nonzero .

From Statement 1, , so the equation becomes

Since the parabola passes through To find , we substitute 0 for and 21 for :

The equation of the parabola is .

Now that the equation is known, the -intercept(s) themselves, if any, can be found by substituting 0 for .