<p>For all x, let the function f be defined by f(x) = a(x-h)^2 + k, where a, h and k are constants. If a and k are positive, which of the following CANNOT be true?</p>
<p>a) f(10) = 1
b) f(0) = -5
c) f(0) = 5
d) f(1) = -h
e) f(-1) = h</p>
<p>I know B can't be true and B is the correct answer, but can someone give me an example for D to prove that D can work?</p>
<p>For any cube, if the volume is V cubic inches and the surface area is A square inches, then V is directly proportional to which if the following?</p>
<p>D can be true because if you let H=-1 and then (x-h)^2=4. Now let A=1/8 and K=1/2. Then f(1)=1=-H. I don’t know if you forgot to say this, but assuming the question said the numbers never had to be integers, test out scenarios where they arent.</p>
<p>Second, </p>
<p>the voulme of the cube = l<em>w</em>H
surface area = 6<em>l</em>w</p>
<p>now in a cube l=w=h so lets just write everything in terms of l</p>
<p>(2) If x is the length of a side of the cube, then V=x^3 and A=6x^2. Solving the second equation for x we get A/6 = x^2, or x = (A/6)^(1/2). So V = x^3 = (A/6)^(3/2) = A^(3/2)/6^(3/2) = kA^(3/2) where k is the constant 1/6^(3/2), choice (E).</p>
<p>Remark: I am assuming that choices (D) and (E) are written incorrectly. They should be:</p>
<p>B cannot be true because f(x) = a(x-h)^2 + k > 0 (because squares are non-negative). D could be true if you let h be a negative number (so that -h is positive). For example, h = -1, a = 1/8, k = 1/2 works as in bllbb6’s explanation.</p>
<p>The second one is a pretty sophisticated problem on “direct variaton.” </p>
<p>Recall that y is directly proportional to x if y=kx for some constant k.</p>
<p>Rspence’s solution is good - a nice quick way to do it on the actual test without getting caught up in what the constant actually is.</p>
<p>I do recommend that you try to understand my solution as well, and in general when practicing try to work out the details in direct and inverse variation problems. This will help you avoid some standard errors.</p>
<p>Let me try to explain each of the following equalities:</p>
<p>V = x^3 = (A/6)^(3/2) = A^(3/2)/6^(3/2) = kA^(3/2) where k is the constant 1/6^(3/2), choice (E).</p>
<p>The first is the definition of volume.</p>
<p>For the second, I replaced x by (A/6)^(1/2) to get [(A/6)^(1/2)]^3 = (A/6)^(3/2). Note here that when you raise a power to a power you multiply the two powers.</p>
<p>For the third I used the rule (x/y)^r = x^r/y^r</p>
<p>And for the fourth I used the fact the x/y = (1/y)*x</p>
<p>Try to write these things on paper - there is an overuse of parentheses here that is unavoidable without a better system for writing equations. It would be nice if this forum would adopt latex or add in an equation editor.</p>