Difference between revisions of "1959 AHSME Problems/Problem 8"
(Created page with "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 <cmath>3...") |
m (improved clarity) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| − | 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 | + | == Problem == |
| + | |||
| + | The value of <math>x^2-6x+13</math> can never be less than: | ||
| + | |||
| + | <math>\textbf{(A)}\ 4 \qquad | ||
| + | \textbf{(B)}\ 4.5 \qquad | ||
| + | \textbf{(C)}\ 5\qquad | ||
| + | \textbf{(D)}\ 7\qquad | ||
| + | \textbf{(E)}\ 13 </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1959|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 11:06, 21 July 2024
Problem
The value of 
can never be less than:
Solution
The 
value at which the minimum value of this quadratic (of the form 
) occurs is 
. The minimum value of the quadratic is therefore at
![\[3^2-6\cdot3+13\]](http://latex.artofproblemsolving.com/9/f/f/9ff27c421764f126f034363088cf568d69bc6d78.png)
![\[=9-18+13\]](http://latex.artofproblemsolving.com/d/c/c/dccb049bcda1242f3df7eb666f123484a24e14e8.png)
![\[=4.\]](http://latex.artofproblemsolving.com/4/e/4/4e4172bd7555609ecc3880f662062bc1edd65993.png)
So, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
