<p>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?</p>
<p>I have no idea how to do this!</p>
<p>what is x-y^2-4, is that mean x=y^2-4 or something? Since there is no “=” and you can never call it a line.</p>
<p>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 L?</p>
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<p>So it gives some points 0,p and 5,t and the equation of the specific graph. So lets put in these points.</p>
<p>0 = y^2 -4, by plugging in the points (0,p), so you get 4 = y^2. this is when you notice something unique. y = ±sqrt(4), so that means the point can either be 0,2 or 0,-2. Let’s keep that in mind and plug in the other point. Using 5,t</p>
<p>5 = y^2 -4 , y^2 = 9, y = ±sqrt(9). So the other point can be either 5,-3 or 5,3. If you connect one of these points with one of the other, you make a line with some slope. It asks for the greatest slope between the 2 points. If you look at the points the most obvious is a positive slope which makes the steepest line, which is made up of the 2 points:</p>
<p>0,-2 and 5,3. Find the slope and it is equal to 1.</p>