In the x-y 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?

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Answer 1 · 2017-11-19T20:53:57

$m = 1$

Given:

Points $(0,p)$ and $(5,t)$ lie on the curve $x = y^2-4$

The slope of a line through the points is:

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

$m = (t-p)/5$

When we solve for t and p, we discover that both have two possible values as follows:

$0 = p^2-4$ and $5 = t^2-4$

$p^2=4$ and $t^2=9$

$p=+-2$ and $t=+-3$

The greatest value will occur when, t is positive and p is negative:

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

$m = 5/5$

$m = 1$

In the x-y coordinate plane, the graph of x=y^2-4x=y^2-4 intersects line l at (0,p)(0,p) and (5,t)(5,t)? What is the greatest possible value of the slope of l?