Question

Consider the graphs of $\left(1\right)y=x^2-\frac{1}{2}x+2$ and $\left(2\right)y=x^2+\frac{1}{2}x+2$ ont he same set of axes. These parabolas have exactly the same shape. Then:

• A. the graphs coincide
• B. the graph of (1) is lower than the graph of (2)
• C. the graph of (1) is to the left of the graph of (2)
• D. the graph of (1) is to the right of the graph of (2)
• E. the graph of (1) is higher than the graph of (2)

Answer 1

Answer: (D)

the graph of (1) is to the right of the graph of (2)

$\left(1\right)⟶y=x^2-\frac{1}{2}x+2⟹y={(x-\frac{1}{4})}^2+\frac{31}{16}\left(2\right)⟶y=x^2+\frac{1}{2}x+2⟹y={(x+\frac{1}{4})}^2+\frac{31}{16}$

Therefore, graph (1) is to the right of graph (2).