DSQ: Calculating the equation of a line - GMAT Quantitative

Find the equation for linear function p(x)

I) and

II)

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

Then, since the x-intercept is known:

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of , which is a parabola. The common -intercept can be found by setting and solving for :

,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

So our equation is as follows

Find the equation for linear function p(x)

I) and

II)

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

Then, since the x-intercept is known:

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of , which is a parabola. The common -intercept can be found by setting and solving for :

,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

So our equation is as follows

Find the equation for linear function p(x)

I) and

II)

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

Then, since the x-intercept is known:

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of , which is a parabola. The common -intercept can be found by setting and solving for :

,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

So our equation is as follows

Find the equation for linear function p(x)

I) and

II)

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

Then, since the x-intercept is known:

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of , which is a parabola. The common -intercept can be found by setting and solving for :

,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.