If  and  be the - and -intercepts, respectively, of a line, the slope of the line is . 

The - and -intercepts of the line are, respectively,  and , so , and consequently, the slope of the green line is .  A line parallel to this line must also have slope . 

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  as its -coefficient is , so this is the correct choice.

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:

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:

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 . If the intercepts are  and , the slope is ; if the intercepts are  and , the slope is . 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, .

We can find out whether the graphs of  and  intersect by first solving for  in the first equation:

We then substitute in the second equation for :

Then we rewrite in standard form:

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 , this is , so here, the discriminant is

.

Therefore, the line of the equation  intersects the parabola of the equation .

We do the same for the other three lines:

Then we rewrite in standard form:

.

The line of  intersects the parabola.

The line of  intersects the parabola.

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 , and it is the correct choice.

The vertex of the parabola of an equation of the form   has -coordinate . Here, we substitute , to obtain -coordinate

.

To find the -coordinate, substitute this for :

The vertex is .

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

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 :

.

If the line includes  and :

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 :

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:

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:

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 . If the intercepts of the line are  and , the slope of the line is ; if the intercepts are  and , the slope is . We choose the latter, since we are looking for a line with positive slope; since its -intercept is , then we can substitute  in the slope-intercept form of the line, , to get the correct equation, .

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:

,

making 9 the correct response.

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 . This makes the slope of the green line . 

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or . Since the desired line must also have -intercept , then the slope-intercept form of the line is

which can be rewritten in standard form:

The vertices of the triangle are the point of intersection of the graphs of the lines of the two equations, and the -intercepts of those lines.

The -intercept of the line of the equation  can be found by setting  and solving for :

The -intercept of the line of the equation  can be found the same way:

The -intercepts are  and ; these are two of the vertices of the triangle, and since the segment connecting them is horizontal, this will be taken as the base. The length of the base is the difference of the -coordinates:

The intersection of the lines of the equations  and  can be found by solving the system of linear equations, as follows:

The point of intersection, and the third vertex of the triangle, is .

Since we are taking the horiztonal segment to be the base, the height will be the vertical distance to this third point - namely, the -coordinate . The area is half the product of the base and the height:

The intercepts of the line of the equation  can be found by substituting 0 for each variable, in turn:

-intercept:

The -intercept is the point 

-intercept:

The -intercept is the point .

This line and the axes together form a right triangle. The horizontal leg is the segment connecting the origin to , and its length is . The vertical leg is the segment connecting the origin to , and its length is . The area of a right triangle is half the product of these legs, which is 

.

Of the five responses, 45 comes closest to the correct area.

If  and  have the same sign, then , and the point has a positive -coordinate. The sign of  is the same as the common sign of  and . Therefore, only the -coordinate restricts the possibilities.

The points witn positive -coordinates are in Quadrant I and Quadrant IV.