<p>In the xy-coordinate plane, the graph of x = y^2 - 4 intersects line L @ (0, p) and (5, t). What is the greatest possible value of the slope of L?</p>
<p>What I did was quickly sketch y = x^2 - 4 (basically ignoring the fact that x and y were reversed, since you can have y be the imput and x the output if you so choose. It didn't say f(x), so I assumed it was ok), and then see what the output values were for 0 and 5 as imput values. 0 Fell at -4 and 5 at 21. Those are 5 apart horizontally and 25 vertically, so my answer was 5. . . . But the answer is 1. I then tried to solve the equation algebraically for y, and got y = sq. root(-x + 4). Since the SAT doesn't deal with imaginary numbers, I'm stumped.</p>
<p>BB page 412, #17
In the xy-coordinate plane, the graph of x = y^2 - 4 intersects line L @ (0, p) and (5, t). What is the greatest possible value of the slope of L?</p>
<p>Sketching a graph put you on the right track but you can't just switch x> and y like that. Plug in some sample values x= y^2 - 4 and y = x^2 - 4 and you'll see your mistake at once.</p>
<p>If you plug some values into the original equation and then graph, you'll see that what you have is a parabola turned sideways starting at (-4,0), two points at (-3,1) and (-3, -1), two points at (0,2) and (0,-2) and so on and so on until you find points at (5, 3) and (5, -3).</p>
<p>The question gives a big clue -- it tells you the graph of x= y^2 - 4 intersects with line L which has points at (0,p) and (5,t). Based on the previous step, you now know that p can equal 2 or -2 and t can equal 3 or -3.</p>
<p>Using the formula for slope (Rise/Run)...
If p = 2 and t = 3, you have a slope of (3-2)/(5-0) = 0.2.
If p = 2 and t = -3, you have a slope of (-3-2)/(5-0) = -1
If p = -2 and t = 3, you have a slope of (3--2)/(5-0) = 1
If p = -2 and t = -3, you have a slope of (-3--2)/(5-0) = -0.2</p>
<p>So, you can see that the largest possible slope for line L is 1. Hope this helps.</p>
<p>the first words, the xy coordinate plane are very important. do not switch the variables in this case. x and y, in this case, are not random variables
Graph the equation (y = square root (x+4). Remember, there is a bottom half to the square root curve, although graphers do not show it. </p>
<p>@ x=0, y can either be + or -2.
@ x = 5, y can either be 3 or -3</p>
<p>For max slope, you want the biggest positive difference between y-values , which are between -2 and +3. Difference between y-values is 5. difference between x-values is 5. slope is 1</p>
<p>p.s. quote : I then tried to solve the equation algebraically for y, and got y = sq. root(-x + 4). Since the SAT doesn't deal with imaginary numbers, I'm stumped.
you made a mistake in calculating. x is positive, and you add 4 to it. (check equation)</p>