The graph of x=y(squared) - 4 intersects line L at (0,p) & (5,t). What's the greatest possible slope of L?

Question

This is a sample SAT question. I have worked out the answer to be 1/5 but the book says the answer is 1. Any explanation?

Answer 1

Note that the intersection means p^2-4 = 0, and t^2-9 = 0. The slope itself is determined by (t-p) / 5, so we need to maximize (t-p). Simple. Let t=3, p=-2. Then (3- -2)/5 = 5/5 = 1.

Steve

Answer 2

So the graph in question is

x = y^2 - 4

and it intersects line L at (0,p) and (5,t)

m = (t - p) / (5 - 0) = (1/5)(t - p)

We know that the points (0,p) and (5,t) are part of the graph
x = y^2 - 4. Thus,

If x = 0, y = p:

0 = p^2 - 4
0 = (p - 2)(p + 2), implying p = {-2, 2}

If x = 5, y = t:

5 = t^2 - 4
0 = t^2 - 9 = (t - 3) (t + 3), implying t = {3, -3}

Since m = (1/5) (t - p), all we have to do is plug in every possibility for t and p and find out what the greatest value is.

p = -2, t = -3

m = (1/5) (-3 - (-2)) = (1/5) (-3 + 2) = (1/5)(-1) = -1/5

p = -2, t = 3

m = (1/5) (3 - (-2)) = (1/5) (3 + 2) = 5/5 = 1

p = 2, t = -3

m = (1/5) (-3 - 2)) = (1/5) (-5) = -1

p = 2, t = 3

m = (1/5) (3 - 2) = (1/5)(1) = 1/5

That means we get the maximum of {-1/5, 1, -1, 1/5}, which is 1.

The greatest possible slope occurs when p = -2, t = 3, and the slope is equal to 1.

Answer 3

The graph is a parabola open to the left.
At x = 0, p = ±2; at x = 5, t = ±3.

The slope of the line is defined as (t-p)/(5-0).

Therefore the maximum slope is obtained when you pick t = 3 and p = -2.

The maximum slope = (3--2)/5 = 1

Answer 4

I assume you got p=2 and t=3, right?

remember that p and t can be negative, too! if you use the points (0,-2) and (5,3), then the answer is 1.