Expert: Sherman D. - 5/16/2005

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?

Answer
x = (y^2) - 4
(0,p) and (5,t)

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

x = (y^2) - 4, (0,p)
0 = p^2 - 4
p^2 = 4
p = ±2

x = (y^2) - 4, (5,t)
t = (5^2) - 4
t = 25 - 4
t = 21

so the intersection points are (0,±2) and (5,21)

using

m = 5/(t - p)
m = 5/(21 - (±2))
m = 5/(21 ± 2)
m = 5/23 or 5/19

the greatest slope value is (5/19)

the formula would be x = (5/19)y - (10/19)

this if (0,p) and (5,t) are still in (x,y) form instead of (y,x) form, because of your equation being x = (y^2) - 4 instead of y = (x^2) - 4.

also because your problem being in x = form instead of y = form, use used m = (x2 - x1)/(y2 - y1) instead of m = (y2 - y1)/(x2 - x1)