Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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== Problem == | == Problem == | ||
− | A [[parabola]] with equation <math> | + | A [[parabola]] with equation <math>y=x^2+bx+c</math> passes through the points <math> (2,3) </math> and <math> (4,3) </math>. What is <math>c</math>? |
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math> | <math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math> | ||
+ | |||
== Solution == | == Solution == | ||
− | Substitute the points (2,3) and (4,3) into the given equation for (x,y). | + | === Solution 1 === |
+ | Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>. | ||
Then we get a system of two equations: | Then we get a system of two equations: | ||
Line 25: | Line 27: | ||
<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer. | <math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer. | ||
− | Alternatively, notice that since the equation is that of a | + | === Solution 2 === |
+ | |||
+ | Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely <math>(3,2)</math>. Thus, the form of the equation of the parabola is <math>y - 2 = (x - 3)^2</math>. Expanding this out, we find that <math>c = 11</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>. | ||
+ | |||
+ | The <math>x</math>-coordinate of the vertex is also equal to <math>\frac{-b}{2a}</math>, so set this equal to <math>3</math> and solve for <math>b</math>, given that <math>a=1</math>: | ||
+ | |||
+ | <math>x=\frac{-b}{2a}</math> | ||
+ | |||
+ | <math>3=\frac{-b}{2}</math> | ||
+ | |||
+ | <math>6=-b</math> | ||
+ | |||
+ | <math>b=-6</math> | ||
+ | |||
+ | Now the equation is of the form <math>y=x^2-6x+c</math>. Now plug in the point <math>(2,3)</math> and solve for <math>c</math>: | ||
+ | |||
+ | <math>y=x^2-6x+c</math> | ||
+ | |||
+ | <math>3=2^2-6(2)+c</math> | ||
+ | |||
+ | <math>3=4-12+c</math> | ||
+ | |||
+ | <math>3=-8+c</math> | ||
+ | |||
+ | <math>\boxed{ \text{(E) }c=11}</math> | ||
+ | |||
+ | === Solution 4 === | ||
+ | Substituting y into the two equations, we get: | ||
+ | |||
+ | <math>3=x^2+bx+c</math> | ||
+ | |||
+ | Which can be written as: | ||
+ | |||
+ | <math>x^2+bx+c-3=0</math> | ||
+ | |||
+ | 4, 2, are the solutions to the quadratic. Thus: | ||
+ | |||
+ | <math>c-3=4\times2</math> | ||
+ | |||
+ | <math>c-3=8</math> | ||
+ | |||
+ | <math>c=11</math> | ||
== See also == | == See also == | ||
Line 31: | Line 78: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 01:38, 6 August 2020
Contents
Problem
A parabola with equation passes through the points and . What is ?
Solution
Solution 1
Substitute the points and into the given equation for .
Then we get a system of two equations:
Subtracting the first equation from the second we have:
Then using in the first equation:
is the answer.
Solution 2
Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely . Thus, the form of the equation of the parabola is . Expanding this out, we find that .
Solution 3
The points given have the same -value, so the vertex lies on the line .
The -coordinate of the vertex is also equal to , so set this equal to and solve for , given that :
Now the equation is of the form . Now plug in the point and solve for :
Solution 4
Substituting y into the two equations, we get:
Which can be written as:
4, 2, are the solutions to the quadratic. Thus:
See also
2006 AMC 10A (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.