A graph of x= y^2 - 4 interesects line L at (0, p) and (5, t). What is the greatest possible slope of L?

Question

A how and why would be nice. Also the answer is not 1/5.

Answer 1

There are two values for p and also for t since the curve has a y^2. It is a parabola that opens to the right and is symetric with respect to the X-axis. The two value of p and t give you four lines (connect each p to two t values giving four lines). Just pick the one with the largest slope.

Slope = (y2 - y1)/(x2 - x1)

(0,p) gives y^2 = 4 or p = +/-2
(5,t) gives y^2 = 9 or t = +/-3

Slopes:
(0,2) to (5,3) = (3 - 2)/(5 - 0) = 1/5
(0,2) to (5,-3) = (-3 - 2)/(5 - 0) = -1
(0,-2) to (5,3) = (3 + 2)/(5 - 0) = 1
(0,-2) to (5,-3) = (-3 + 2)/(5 - 0) = -1/5

So the largest slope is 1.

Answer 2

m = 6/5


0 = y^2 -4, y = 2, -2
5 = t^2 -4, t = 3, -3

4 lines can be drawn using 4 points (0, 2) (0, -2) (5, 3) (5, 3)

Line with larger slope appears to be (0, 2) (5, -3) or (0, -2) (5, 3)

Answer 3

(0,p)

x = y^2-4
0 = p^2- 4
p = ±2

(5,t)

x = y^2-4
5 = t^2- 4
t = ±3



Slope of L = change in y / change in x

Possible coordinates of intersection:
(0,-2) or (0,2)
and
(5,-3) or (5,3)

Possible values of slope are
-1/5, 1, -1, 1/5

The greatest possible slope of L would be 1 between the points:
(0, -2) and (5, 3)

Answer 4

y= mx + b......plug and chug those values you are given for x my friend