1954 AMC 12 Problem 42

Question

Consider the graphs of

(1)

y=x^2-\dfrac12x+2

and

(2)

y=x^2+\dfrac12x+2

on the same set of axes. These parabolas have exactly the same shape. Then:

the graphs coincide.

the graph of

(1)

is lower than the graph of

(2).

the graph of

(1)

is to the left of the graph of

(2).

the graph of

(1)

is to the right of the graph of

(2).

the graph of

(1)

is higher than the graph of

(2).

Find the

x

-coordinate of each vertex using

-\frac{b}{2a}

Graph

(1)

has vertex

x=\frac{1}{4}

, whereas graph

(2)

has vertex

x=-\frac{1}{4}

The vertex of

y=x^2+bx+2

has

x

-coordinate

-\frac{b}{2}.

Thus graph

(1)

has its vertex at

x=\frac{1}{4},

while graph

(2)

has its vertex at

x=-\frac{1}{4}.

Their vertex heights are equal, so graph

(1)

is the same parabola shifted to the right.

Thus, the correct answer is D.

Problem 42 in other years