Coordinate Geometry

[gusterguy]

Hi guys,

So here's the question: In the xy-coordinate plane, the graph of x = y^2 - 4 intersects line l at (0, p) and (5, t). What is the greatest possible value of the slope of l?

So i'm an SAT tutor and have been for months now - I think I do a very good job in all of the subjects and love working with high schoolers. It wasn't until yesterday, there were a few math problems that I knew in theory how to do but couldn't arrive at the right answer. Hopefully you guys can help me and I will in turn rely the information to my students.

So for the problem above, I realized that since y is squared that the y value that can work for the two points will be either positive or negative. So I ended up with the points (0, 2), (0, -2), (5, 3), and (5,-3). The answer for the greatest possible value is 1 - I wondered besides finding the slopes for the 4 various combinations of pairs of points - is there any easier way to arrive at the answer? Also when I see a variable squared - should I always be on the lookout that positive and negative values will work for it?

Thanks!

 

[stapel]

In the xy-coordinate plane, the graph of x = y^2 - 4 intersects line l at (0, p) and (5, t). What is the greatest possible value of the slope of l?

Draw a quick graph (it doesn't have to be "accurate") of x = y[sup:2w901m20]2[/sup:2w901m20] - 4, marking off the intercepts at (-4, 0), (0, -2), and (0, 2). Draw the vertical lines x = 0 and x = 5, noting that (0, p) lies somewhere on the former and (5, t) lies somewhere on the latter.

Each of these vertical lines cross the sideways parabola twice, giving you four choices for the line \(\displaystyle l\). It is pretty obvious which one of these will have the biggest slope. :wink:

You already have the coordinates of the lower-left-hand point, and the upper-right-hand point is easily found. Then plug-n-chug to find the slope, and you're done!

Eliz.

 

[PAULK]

In the xy-coordinate plane, the graph of x = y^2 - 4 intersects line l at (0, p) and (5, t). What is the greatest possible value of the slope of l?

If x = y^2 - 4 p.t. (0,p), then p = +- 2,

If x = y^2 - 4 p.t. (5,t), then t = +- 3,

So your four candidate points are:

A(0,2) or B(0,-2)
and
C(5,3) or D(5,-3)

The largest slope would be one of these four:

m(AD) = 5/(-5) = -1
m(AC) = 5(1) = 5 << that's it.
m(BC) = 5/5 = 1
m(BD) = 5/(-1) = -5.

 

[stapel]

If x = y^2 - 4 p.t. (0,p), then p = +- 2,

If x = y^2 - 4 p.t. (5,t), then t = +- 3

If, as the original poster explained, you're wanting to teach this to students, then you may want to consider going into a bit more detail. (Since the poster already arrived at the answer, and was requesting a simpler or faster solution method, I am assuming that you are outlining how you would teach this material.)

So your four candidate points are: A(0,2) or B(0,-2), and C(5,3) or D(5,-3). The largest slope would be one of these four:

m(AD) = 5/(-5) = -1
m(AC) = 5(1) = 5 << that's it.
m(BC) = 5/5 = 1
m(BD) = 5/(-1) = -5.

Actually, the values are as follows:

. . . . .\(\displaystyle m_{AD}\, =\, \frac{2\, +\, 3}{0\, -\, 5}\, =\, \frac{5}{-5}\, = -1\)

. . . . .\(\displaystyle m_{AC}\, =\, \frac{2\, -\, 3}{0\, -\, 5}\, =\, \frac{-1}{-5}\, = \frac{1}{5}\)

. . . . .\(\displaystyle m_{BC}\, =\, \frac{-2\, -\, 3}{0\, -\, 5}\, =\, \frac{-5}{-5}\, = 1\)

. . . . .\(\displaystyle m_{BD}\, =\, \frac{-2\, +\, 3}{0\, -\, 5}\, =\, \frac{1}{-5}\, = -\frac{1}{5}\)

So the solution is actually the one that the original poster had displayed. :shock:

Eliz.