1955 AHSME Problems/Problem 28

Problem 28

On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation. If $b \neq 0$ and $c \neq 0$ these two graphs intersect:

$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis}$

Solution

Replacing $x$ with $-x,$ we have the graph of $y = a(-x)^2 + b(-x) + c = ax^2 - bx+c.$

We can plug in simple values of $a, b,$ and $c$ for convenient drawing. For example, we can graph $x^2+2x+1$ and $x^2-2x+1.$ We see that the parabolas intersect at $(0, 1),$ which is on the y-axis.

The answer is $\boxed{\textbf{(E)}}.$

1955 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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.

Page

Toolbox

Search

1955 AHSME Problems/Problem 28

Problem 28

Solution

See Also