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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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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>
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So our answer is <math>\boxed{C}</math>
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So our answer is <math>\fbox{\textbf{(C)}}</math>.
  
 
~JustinLee2017
 
~JustinLee2017
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== See Also ==
 +
{{AHSME 50p box|year=1957|num-b=9|num-a=11}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 09:09, 25 July 2024

Solution

This graph generates a parabola, since the degree of $x$ is $2$. The $x-$ coordinate of the vertex of a parabola given by $ax^2 + bx + c$ is at $\frac{-b}{2a}$ So, the vertex of this parabola is at \[x = \frac{-4}{2(2)} = \frac{-4}{4} = -1\] Since the coefficient of $x^2$ is positive, at $x = -1$, the parabola is at its minimum. Substituting $x = -1$, we get \[y = 2(-1)^2 + 4(-1) + 3 \Rrightarrow y = 2 -4 + 3 \Rrightarrow y = 1\] So our answer is $\fbox{\textbf{(C)}}$.

~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

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