Difference between revisions of "1998 AHSME Problems/Problem 14"
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− | == Problem | + | == 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? | ||
<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> | ||
− | + | ==Solution== | |
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Thus <math>\boxed{A}</math> is the right answer - only <math>a</math> is positive. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1998|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:29, 5 July 2013
Problem
A parabola has vertex of and has two
intercepts, one positive, and one negative. If this parabola is the graph of
which of
and
must be positive?
Solution
The vertex of the parabola is at . Since there are two x-intercepts, it must open upwards. If it opened downard, there would be no roots. Thus,
.
The x-coordinate of the vertex is . Since
is positive, and the x-intercept is positive, the value
must be positive too, and
is negative.
By Vieta, the product of the two roots is . Since the two roots are a positive number and a negative number, the product is negative. Since
is positive, that means
must be negative.
Thus is the right answer - only
is positive.
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |
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