<p>Following is a math problem that looks quite difficult. I'm totally not sure how to go about solving it. I've never encountered such prob before. Help me understand it, guys :)</p>
<p>15) 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>Many thanks in advance :)</p>
<p>so the answer is B
f(0) = a(-h)^2+k
f(0) = ah^2+k
-5=ah^2+k
the question states that a and k are positive also h^2 is positive… so it is impossible to give a negative answer.</p>
<p>It just takes a few seconds if you recognize that f(x)=a(x-h)^2+k is the equation of a parabola with its vertex at the point (h,k).</p>
<p>Because a is positive, the parabola opens upward.</p>
<p>Because k is positive, the vertex is above the x-axis.</p>
<p>And the parabola goes up from there on both sides, so the y-coordinates of points on the parabola will never be negative.</p>
<p>If you don’t want to think about it graphically (which is what I’m sure the authors of this question intend you to do), Nader’s numerical analysis is also correct.</p>
<p>Can someone disprove choice D please?</p>
<p>
</p>
<p>a= 1/9
k= 1
h= -2</p>
<p>when x=1
f(x)=2</p>
<p>-h= a(1-h)^2 + k = a(1-2h+h^2)+k=a-2ha+h^2+k. If h is a positive number, then -2ha is negative because h and a are positive. So if h and a are big numbers, it is possible for a-2ha+h^2 + k to equal a negative number (-h) if say k is 1, and a and h are fairly big numbers or even fractions.</p>
<p>Thank you so much all :)</p>