2006 Cyprus MO/Lyceum/Problem 4

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 $f$ ?

$\mathrm{(A)}\ \text{intersects x-axis}\qquad\mathrm{(B)}\ \text{touches y-axis}\qquad\mathrm{(C)}\ \text{touches x-axis}\qquad\mathrm{(D)}\ \text{has minimum point}\qquad\mathrm{(E)}\ \text{has maximum point}$

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 $\mathrm{(C)}$ .

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 3	Followed by Problem 5
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2006 Cyprus MO/Lyceum/Problem 4

Problem

Solution

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