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

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== Problem 14 ==
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== Problem ==
 
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^2 + bx + c,</math> which of <math>a,b,</math> and <math>c</math> must be positive?
 
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^2 + bx + c,</math> which of <math>a,b,</math> and <math>c</math> must be positive?
  
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Thus <math>\boxed{A}</math> is the right answer - only <math>a</math> is positive.
 
Thus <math>\boxed{A}</math> is the right answer - only <math>a</math> is positive.
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== See also ==
 +
{{AHSME box|year=1998|num-b=12|num-a=14}}

Revision as of 22:37, 7 August 2011

Problem

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.

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions