Difference between revisions of "2005 AMC 12B Problems/Problem 8"

(Problem)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
For how many values of <math>a</math> is it true that the line <math>y = x + a</math> passes through the
 +
vertex of the parabola <math>y = x^2 + a^2</math> ?
 +
 +
<math>
 +
\mathrm{(A)}\ 0      \qquad
 +
\mathrm{(B)}\ 1      \qquad
 +
\mathrm{(C)}\ 2      \qquad
 +
\mathrm{(D)}\ 10      \qquad
 +
\mathrm{(E)}\ \text{infinitely many} 
 +
</math>
  
 
== Solution ==
 
== Solution ==
 
+
We see that the vertex of the quadratic function <math>y = x^2 + a^2</math> is <math>(0,\,a^2)</math>. The y-intercept of the line <math>y = x + a</math> is <math>(0,\,a)</math>. We want to find the values (if any) such that <math>a=a^2</math>. Solving for <math>a</math>, the only values that satisfy this are <math>0</math> and <math>1</math>, so the answer is <math>\boxed{\mathrm{(C)}\ 2}</math>
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
+
{{AMC12 box|year=2005|ab=B|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 17:23, 9 September 2020

Problem

For how many values of $a$ is it true that the line $y = x + a$ passes through the vertex of the parabola $y = x^2 + a^2$ ?

$\mathrm{(A)}\ 0      \qquad \mathrm{(B)}\ 1      \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 10      \qquad \mathrm{(E)}\ \text{infinitely many}$

Solution

We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$. The y-intercept of the line $y = x + a$ is $(0,\,a)$. We want to find the values (if any) such that $a=a^2$. Solving for $a$, the only values that satisfy this are $0$ and $1$, so the answer is $\boxed{\mathrm{(C)}\ 2}$

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png