14 years ago
frnkcharles

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

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?
Top 4 Answers
14 years ago
Puggy
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.
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14 years ago
sahsjing
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
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14 years ago
John C
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.
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