Difference between revisions of "1957 AHSME Problems/Problem 10"
(Created page with "==Solution== This graph generates a parabola, since the degree of <math>x</math> is <math>2</math>. The <math>x-</math> coordinate of the vertex of a parabola given by <math>a...") |
m (see also box) |
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Line 6: | Line 6: | ||
Substituting <math>x = -1</math>, we get | Substituting <math>x = -1</math>, we get | ||
<cmath>y = 2(-1)^2 + 4(-1) + 3 \Rrightarrow y = 2 -4 + 3 \Rrightarrow y = 1</cmath> | <cmath>y = 2(-1)^2 + 4(-1) + 3 \Rrightarrow y = 2 -4 + 3 \Rrightarrow y = 1</cmath> | ||
− | So our answer is <math>\ | + | So our answer is <math>\fbox{\textbf{(C)}}</math>. |
~JustinLee2017 | ~JustinLee2017 | ||
+ | |||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 50p box|year=1957|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 09:09, 25 July 2024
Solution
This graph generates a parabola, since the degree of is
. The
coordinate of the vertex of a parabola given by
is at
So, the vertex of this parabola is at
Since the coefficient of
is positive, at
, the parabola is at its minimum.
Substituting
, we get
So our answer is
.
~JustinLee2017
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.