<p>Please refer to questions :
Q 14 and 15 (pg 340)</p>
<p>Q-14 why is II right?
Q-15 how doesit go with the parabola curve in squared eqns.</p>
<p>Q 6 (pg 369)</p>
<p>doesn't the graph shift 5 units to the left..if that happens nothing will will intersect between 0-12?</p>
<p>Please provide reasonings for your answer. i just could not understand the logic behind it.
thanks..</p>
<p>p340,14, since a>b,so a-b>0; a(a-b)=0,a doesn’t equally to b, thus a must be 0, a-b can’t be 0, b<a, b is negative. Thus three answers are all right.
15, put (-2,0) into the following functions and see which one is right.That’s enough.
p369, there is no no.6 on this page.i suppose it should be no.8.</p>
<p>It’s telling you that the function repeats every 5 units. For example, f(11) = f(6 +5) = f(6) = f(1 + 5) = f(1). f(x) =0 only 4 times between [0,5], so it is zero 8 times between [0,10]. From [10,12], the graph is the same as [0,2], so there is 1 more zero for 9 zeros in total.</p>
<p>14) a=0, and a>b so 0>b</p>
<p>15) y = x^2 is an upward parabola and x = y^2 is a rightward parabola
you switch the x and the y… so it flips around (inverse graph)…you can also graph the answer choices and see which one fits the picture to find the answer</p>
<p>8)
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mistake on 3rd graph, there should only be the first red mark</p>
<p>f(x+5) = f(x). which means when you shift it 5 to the left the graph will look the same. so every x on this graph repeats itself on f(x+5).</p>
<p>notice how the graph ends the same way it starts? you can see that after x hits 5, the graph starts repeating. count that from x=0 to x=5 the graph hits y = 0 (the x-axis) 4 times. Since it repeats itself again at f(x+5), or from x=5 to x=10, the graph hits y = 0 another 4 times. From x=10 to x=12 the graph repeats the same as x=0 to x=2, which intersects another time, totaling 9 times (4+4+1)</p>
