Graphing a quadratic function - GMAT Quantitative

If the graph of is concave upward, then .

If the graph does not intersect the -axis, then has no real solution, and the discriminant is negative:

For the parabola to have both characteristics, it must be true that and , but these two events are mutually exclusive. Therefore, the parabola cannot exist.

The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

The line of symmetry of a vertical parabola is a vertical line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

A parabola with its -intercepts at and at has as its equation

for some nonzero . If this is multiplied out, the equation can be rewritten as

or, simplified,

The sign of quadratic coefficient determines whether it is concave upward or concave downward. We do not have the sign or any way of determining it.

The -coordinate of the -intercept is the contant, , but without knowing , we have no way of knowing .

The -coordinate of the vertex of is the value . since , this expression becomes

The -coordinate is

,

but without knowing , this coordinate, and the vertex itself, cannot be determined.

The line of symmetry is the line ; this value was computed to be equal to 6, so the line can be determined to be .

This can be answered rewriting this expression in the form

.

Once this is done, we can identify the vertex as the point .

The vertex is

I) The orientation of the parabola is determined solely by the value of . Since , the parabola can be determined to open upward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I only.

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

I) The orientation of the parabola is determined solely by the sign of . Since , a positive value, the parabola can be determined to be concave upward.

II) The -coordinate of the vertex is ; since you given both and , you can find this to be

The -coordinate is equal to . However, you need the entire equation to determine this value; since you do not know , you cannot find the -coordinate. Therefore, you cannot find the vertex.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and is unknown, the -intercept(s) cannot be identified.

V) The line of symmetry has equation . When exploring the vertex, we found that this value is equal to , so the line of symmetry is the line of the equation .

The correct response is I and V only.

I) The orientation of the parabola is determined solely by the sign of . It is given in the problem that is negative, so it follows that the parabola is concave downward.

II) The -coordinate of the vertex is ; since , this number is . The -coordinate is , but since we do not know the values of , , and , we cannot find this value. Therefore, we cannot know the vertex.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since , this can be rewritten and simplified as follows:

However, since has no real square root, has no real solutions, and its graph has no -intercepts.

V) The line of symmetry has equation . When exploring the vertex, we found that this value is equal to , so the line of symmetry is the line of the equation .

The correct response is I, IV, and V only.

The graph of has as its line of symmetry the vertical line of the equation

Since in each choice, we want to find such that

so the correct choice is .

The graph of has exactly one -intercept if and only if

has exactly one solution - or equivalently, if and only if

Since in all three equations, , we find the value of that makes this statement true by substituting and solving:

The correct choice is .

A horizontal parabola has as its equation, in standard form,

,

with real, nonzero.

Its orientation depends on the sign of . In the equation of a concave-right parabola, is positive, so the correct choice is .

The -coordinate of the -intercept of the graph of a function of the form - a quadratic function - is the point . Since , the -intercept is at the point .

Because of this, the graph of has its -intercept at . Among the other choices, only has a graph with its -intercept also at .

The eqiatopm of is given in the vertex form

,

so the vertex of its parabola is . The graphs of and are parabolas with the same vertex, so they must have the same values for and .

For the function , and .

Of the five choices, the only equation of that has these same values, and that therefore has a parabola with the same vertex, is .

To verify, graph both functions on the same grid.

The system of equations can be rewritten as

.

We can set the two expressions in equal to each other and solve:

We can substitute back into the equation , and see that either or . The latter value is the correct choice.

The -intercepts of the parabola

can be determined by setting and solving for :

or

The intercepts of the parabola are and .

We can check each equation to see whether these two ordered pairs satisfy them.

:

Each of these equations has a parabola that does not have as an -intercept. However, we look at

and are also -intercepts of the graph of this function, so this is the correct choice.

We can set the two quadratic expressions equal to each other and solve for .

and , so

The -coordinates of the points of intersection are 2 and 6. To find the -coordinates, substitute in either equation:

One point of intersection is .

The other point of intersection is .

1 is not among the choices, but 41 is, so this is the correct response.

The graph of a function of the form - a quadratic function - is a vertical parabola with line of symmetry .

The graph of the function therefore has line of symmetry

, or

We examine all four definitions of to find one with this line of symmetry.

:

, or

:

, or

, or

, or

Since the graph of the function has the same line of symmetry as that of the function , that is the correct choice.

The -intercept of the graph of , a parabola, can be found by setting and solving for :

The -intercept is .

The -intercepts can be found by and solving for :

, in which case , or

, in which case .

The two -intercepts are and .

The triangle is shown below:

If the segment connecting the points on the -axes is taken as the base, then

The height is therefore the vertical distance from this segment to the point on the -axis, which is

The area is half their product, or

.

The vertex of the graph of a quadratic equation can be found by setting , substituting, and solving for . Setting , this is

Substitute:

The vertex is the point .

The -intercepts can be found by and solving for :

, in which case , or

, in which case .

The -intercepts are and .

The triangle with vertices , , and is below:

If we take the base to be the horizontal segment connecting the -intercepts, then

.

The height is therefore the vertical distance from this segment to the vertex, which is

The area of the triangle is half their product:

.