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Difference between revisions of "1998 AHSME Problems/Problem 14"

(Created page with "== Problem 14 == A parabola has vertex of <math>(4,-5)</math> and has two <math>x-</math>intercepts, one positive, and one negative. If this parabola is the graph of <math>y = ax...")
 
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<math> \mathrm{(A) \ } \text{only}\ a \qquad \mathrm{(B) \ } \text{only}\ b \qquad \mathrm{(C) \ } \text{only}\ c \qquad \mathrm{(D) \ } a\ \text{and}\ b\ \text{only} \qquad \mathrm{(E) \ } \text{none}</math>
 
<math> \mathrm{(A) \ } \text{only}\ a \qquad \mathrm{(B) \ } \text{only}\ b \qquad \mathrm{(C) \ } \text{only}\ c \qquad \mathrm{(D) \ } a\ \text{and}\ b\ \text{only} \qquad \mathrm{(E) \ } \text{none}</math>
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==Solution==
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The vertex of the parabola is at <math>(4,-5)</math>.  Since there are two x-intercepts, it must open upwards.  If it opened downard, there would be no roots.  Thus, <math>a > 0</math>.
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The x-coordinate of the vertex is <math>\frac{-b}{2a}</math>.  Since <math>a</math> is positive, and the x-intercept is positive, the value <math>-b</math> must be positive too, and <math>b</math> is negative.
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By Vieta, the product of the two roots is <math>\frac{c}{a}</math>.  Since the two roots are a positive number and a negative number, the product is negative.  Since <math>a</math> is positive, that means <math>c</math> must be negative.
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Thus <math>\boxed{A}</math> is the right answer - only <math>a</math> is positive.
  
 
[[1998 AHSME Problems/Problem 14|Solution]]
 
[[1998 AHSME Problems/Problem 14|Solution]]

Revision as of 21:36, 7 August 2011

Problem 14

A parabola has vertex of $(4,-5)$ and has two $x-$intercepts, one positive, and one negative. If this parabola is the graph of $y = ax^2 + bx + c,$ which of $a,b,$ and $c$ must be positive?

$\mathrm{(A) \ } \text{only}\ a \qquad \mathrm{(B) \ } \text{only}\ b \qquad \mathrm{(C) \ } \text{only}\ c \qquad \mathrm{(D) \ } a\ \text{and}\ b\ \text{only} \qquad \mathrm{(E) \ } \text{none}$

Solution

The vertex of the parabola is at $(4,-5)$. Since there are two x-intercepts, it must open upwards. If it opened downard, there would be no roots. Thus, $a > 0$.

The x-coordinate of the vertex is $\frac{-b}{2a}$. Since $a$ is positive, and the x-intercept is positive, the value $-b$ must be positive too, and $b$ is negative.

By Vieta, the product of the two roots is $\frac{c}{a}$. Since the two roots are a positive number and a negative number, the product is negative. Since $a$ is positive, that means $c$ must be negative.

Thus $\boxed{A}$ is the right answer - only $a$ is positive.

Solution

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