Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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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>? | 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> \ | + | <math> \textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11 </math> |
− | + | == Solution 1 == | |
− | |||
− | |||
Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>. | Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>. | ||
Line 25: | Line 23: | ||
<math>0=1+-12+c</math> | <math>0=1+-12+c</math> | ||
− | <math>c= | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
− | === Solution | + | === Solution 1.1 === |
− | Alternatively, notice that since the equation is that of a | + | 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=\boxed{\textbf{(E) }11}</math>. |
− | + | == Solution 2 == | |
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 points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>. | ||
Line 55: | Line 53: | ||
<math>3=-8+c</math> | <math>3=-8+c</math> | ||
− | <math>\boxed{c= | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
+ | |||
+ | == Solution 3 == | ||
+ | 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> | ||
+ | |||
+ | <math>4</math> and <math>2</math> are the solutions to the quadratic. Thus: | ||
+ | |||
+ | <math>c-3=4\times2</math> | ||
+ | |||
+ | <math>c-3=8</math> | ||
+ | |||
+ | <math>c=\boxed{\textbf{(E) }11}</math>. | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:10, 16 December 2021
Problem
A parabola with equation passes through the points
and
. What is
?
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:
.
Solution 1.1
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 2
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 3
Substituting y into the two equations, we get:
Which can be written as:
and
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.