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Difference between revisions of "1959 AHSME Problems/Problem 8"

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== Solution ==  
 
== Solution ==  
  
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The <math>x</math> value at which the minimum value of this quadratic occurs is <math>-\frac{-6}{2\cdot1}=3</math>. The minimum value of the quadratic is therefore at
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The <math>x</math> value at which the minimum value of this [[quadratic]] (of the form <math>ax^2+bx+c</math>) occurs is <math>\frac{-b}{2a}=-\frac{-6}{2\cdot1}=3</math>. The minimum value of the quadratic is therefore at
 
<cmath>3^2-6\cdot3+13</cmath>
 
<cmath>3^2-6\cdot3+13</cmath>
 
<cmath>=9-18+13</cmath>
 
<cmath>=9-18+13</cmath>
 
<cmath>=4.</cmath>
 
<cmath>=4.</cmath>
 
So, the answer is <math>\boxed{\textbf{(A)} \ 4}</math>.
 
So, the answer is <math>\boxed{\textbf{(A)} \ 4}</math>.
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 +
==See also==
 +
{{AHSME 50p box|year=1959|num-b=7|num-a=9}}
 +
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 11:06, 21 July 2024

Problem

The value of $x^2-6x+13$ can never be less than:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 13$

Solution

The $x$ value at which the minimum value of this quadratic (of the form $ax^2+bx+c$) occurs is $\frac{-b}{2a}=-\frac{-6}{2\cdot1}=3$. The minimum value of the quadratic is therefore at \[3^2-6\cdot3+13\] \[=9-18+13\] \[=4.\] So, the answer is $\boxed{\textbf{(A)} \ 4}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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