<p><a href="http://i43.tinypic.com/m95e8z.jpg%5B/url%5D">http://i43.tinypic.com/m95e8z.jpg</a>
<a href="http://i43.tinypic.com/350kn7q.jpg%5B/url%5D">http://i43.tinypic.com/350kn7q.jpg</a></p>
<p>Sorry that they're sideways, I was too lazy to put them in the right position. When I read these kind of math questions I think that they're easy to do but, the only thing is I don't know how to work them out..</p>
<p>For the first question, I found the easiest solution was to substitute values for numbers, and work my way from there. Unless you like algebra.</p>
<p>For the second one, you need to know what it means when f(x) = x is changed to f(x) = (x+h) or f(x) = x + h</p>
<p>When you add or subtract a number from X itself, like f(x)=(x-2)+5(x-2), that will mean either it moves a number of units to the left or right. For example, f(x) = x-2 means that the entire graph will move two units to the right. On the other hand, f(x) = x+2 means that the entire graph will move two units to the left.</p>
<p>Same concept with adding a number to the entire equation of x and not x itself. For example, f(x) = (x^2 + 5x) + 2. That means the entire graph will move either up or down. In this case it will move two units up. However, if it was f(x) = (x^2 + 5x) - 2, then it would move two units down.</p>
<p>Let’s apply the previous concepts to our question. The point (-1,3) was changed into (2,1). That means the graph moved 2 units down (the value of k is -2) and it moved 3 units to the right (the value of h is -3).</p>
<p>Therefore, hk = -3 x -2 = 6</p>
<p>Since x^(-4/3)=k^(-2), you can raise both sides to 1/2 in order to bring x to be raised to x^(-2/3) The equation is now x^(-2/3) = k^(-1).</p>
<p>y^(4/3) =n^2; raise both sides by -1/2.
y^(-2/3) = n^(-1)</p>
<p>Since (xy)^(-2/3) = x^(-2/3) * y^(-2/3), you can substitute the derived equations into each variable.</p>
<p>k^(-1) * n^(-1) = 1/nk</p>
