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 $(2).$

Answer: D

Concepts:parabola completing the square transformation

Difficulty rating: 1400

Small Hint:

Find the $x$ -coordinate of each vertex using $-\frac{b}{2a}$

Big Hint:

Graph $(1)$ has vertex $x=\frac{1}{4}$ , whereas graph $(2)$ has vertex $x=-\frac{1}{4}$

Solution:

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.

1954 AMC 12 Problem 42

Answer: D

Difficulty rating: 1400

Small Hint:

Big Hint:

Solution:

Problem 42 in Other Years