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 $f$ ?

A. intersects x-axis

B. touches y-axis

C. touches x-axis

D. has minimum point

E. 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 with the x-axis (it is tangent to it). $\mathrm{(B)}$ is not true; the graph intersects with 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.

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

See also