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2006 Cyprus MO/Lyceum/Problem 4

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Problem

Given the function f(x)=\alpha x^2 +9x+ \frac{81}{4\alpha} , \alpha \neq 0 Which of the following is correct, about the graph of f?

\mathrm{(A)}\ \text{intersects x-axis}\qquad\mathrm{(B)}\ \text{touches y-axis}\qquad\mathrm{(C)}\ \text{touches x-axis}\qqua...

Solution

\alpha x^2+9x+\frac{81}{4\alpha}=\left(\sqrt{\alpha}x+\frac{9}{2\sqrt{\alpha}}\right)^2 Notice that if f(x) = 0, then x has the unique root of -\frac{\frac{9}{2\sqrt{\alpha}}}{\sqrt{\alpha}} = \frac{-9}{2\alpha}, so it touches the x-axis, \mathrm{(C)}.

From above, \mathrm{(A)} is not correct because the graph does not intersect the x-axis (it is tangent to it).

\mathrm{(B)} is not true; the graph intersects the y-axis since the parabola opens up or down.

\mathrm{(D)} and \mathrm{(E)} depend upon the value of \alpha; if \alpha > 0, then the parabola has a minimum, and if \alpha > 0 then the parabola has a maximum.

Thus, the answer is \mathrm{(C)}.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 3
Followed by
Problem 5
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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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