SAT Question?

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?

Somebody explain to me like I'm a 2 year old how you get this.
(btw, the answer key says its 1.)

Answer 1

Plug in the x-values in x = y² - 4

(0, p)
0 = p² - 4
4 = p²
p = 2 or -2

(5, t)
5 = t² - 4
9 = t²
t = 3 or -3

So the possible points are:

(0, 2), (5, 3),
(0, 2), (5, -3),
(0, -2), (5, 3), or
(0, -2), (5, -3).

The slope is given by:

(y2 - y1)/(x2 - x1)

So choose the y-values that are the farthest apart: 2 and -3 or 3 and -2 to maximize the slope.

(3 - -2)/(5 - 0) = 5/5 = 1 or
(2 - -3)/(5 - 0) = 5/5 = 1

Either way, you get 1.

Answer 2

The equation of the parabola is

x = y² - 4

The two point of intersection of line L with the parabola are

(0,p) and (5,t)

The slope m, of the line thru the two points is

m = ∆y/∆x = (t - p)/(5 - 0) = (t - p)/5

Now let's solve for t and p.

First we need to recast the equation of the parabola.

x = y² - 4
y² = x + 4
y = ±√(x + 4)

Solve for t.

y = ±√(5 + 4) = ±√9 = ±3

Solve for p.

y = ±√(0 + 4) = ±√4 = ±2

To maximize m, maximize ∆y by making t positive and p negative (or the other way around--it doesn't make any difference).

Then
t = 3
p = -2

∆y = t - p = 3 - (-2) = 5

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

Answer 3

slope(m) = (t-p)/5
But
0 = p^2-4,
p = ±2

5 = t^2-4
t = ±3

t = 3 and p = -2 will give you the greatest possible value of the slope of L.
slope(m) = (3+2)/5 = 1

Answer 4

plug in 0 and 5 for X into function x = (y^2)-4 to get the Y value when X = 0 and X = 5.
Then use slope formula to find slope.
SAT questions are meant to be simple - you can usually solve them in a few steps even when they seem complicated.