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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 4"

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==Problem==
 
==Problem==
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{{empty}}
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Given the function <math>f(x)=\alpha x^2 +9x+ \frac{81}{4\alpha}</math> , <math>\alpha \neq 0</math>
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Which of the following is correct, about the graph of <math>f</math>?
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<math>\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}</math>
  
 
==Solution==
 
==Solution==
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{{solution}}
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<cmath>
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\alpha x^2+9x+\frac{81}{4\alpha}=\left(\sqrt{\alpha}x+\frac{9}{2\sqrt{\alpha}}\right)^2
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</cmath>
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Notice that if <math>f(x) = 0</math>, then <math>x</math> has the unique root of <math>-\frac{\frac{9}{2\sqrt{\alpha}}}{\sqrt{\alpha}} = \frac{-9}{2\alpha}</math>, so it touches the x-axis, <math>\mathrm{(C)}</math>.
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From above, <math>\mathrm{(A)}</math> is not correct because the graph does not intersect the x-axis (it is tangent to it).
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<math>\mathrm{(B)}</math> is not true; the graph intersects the y-axis since the [[parabola]] opens up or down.
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<math>\mathrm{(D)}</math> and <math>\mathrm{(E)}</math> depend upon the value of <math>\alpha</math>; if <math>\alpha > 0</math>, then the parabola has a minimum, and if <math>\alpha > 0</math> then the parabola has a maximum.
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Thus, the answer is <math>\mathrm{(C)}</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=3|num-a=5}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=3|num-a=5}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 10:46, 27 April 2008

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 $\mathrm{(C)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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