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 .

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 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 -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.

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 .

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 .

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 -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 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:

.