<p>okay so the above link is the graph and heres the question.</p>
<p>Which of the following could be the equation of the function graphed in the xy-plane above?</p>
<p>A y = (x)^2 + 1
B y = x^2 + 1
C y = |x^2 + 1|
D y = |x^2 1|
E y = |(x 1)^2|</p>
<p>The answer is D. Why?</p>
<p>there must be an absolute value sign, so A, B are both out.
Now if you plug in x=1, you get y=0, C is out.
Now plug in x=-1, y=0, only D works.</p>
<p>Or…</p>
<p>You should kinda realize that this is a parabola, and that an absolute value sign is needed to make that middle portion positive (as it should be in the third and fourth quadrants if no absolute value sign existed). So, you should envision a graph of a parabola which bottoms out at (0,-y). Because the graph isn’t scaled, y can pretty much be anything as of now. </p>
<p>So, looking at the answer choices, A and B don’t work because there is no absolute value sign. C is wrong because, without the absolute value sign, the y-int is at (0,1), where the y value is positive. E doesn’t work because this parabola’s shift will shift the entire parabola 1 unit to the right (-1 is inside the parenthesis, therefore affects x), which doesn’t get the desired y-int w/o the abs. value sign but instead gets (0,1) again. Therefore, D is the answer, and it’s y-int w/o the abs. value sign is (0,-1) which is the only equation to satisfy that condition. </p>
<p>If you’re having trouble with this problem, I suggest you look into absolute value and graph transformations.</p>
<p>And my method may seem long, but for me this problem took like 5-10 seconds to solve with said method. I’m just writing out the entire ideal thought process to help you better, although I skipped some of the above steps in the actual problem. My point is: Don’t necessarily shirk the algebraic way while instead learning how to properly guess and check or plug in numbers or whatever.</p>
