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. | ||
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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. | ||
| + | |||
| + | 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 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
