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- anonymous

In the standard form of the absolute value function, f(x) = a|x − h| + k, what is the significance of coefficient a?
(A)It determines which point will be the vertex of the graph.
(B)It determines whether the graph is V-shaped, U-shaped, or a straight line.
(C)It determines the distance of the vertex from the x-axis.
(D)It determines the width of the graph.

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- anonymous

- jamiebookeater

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- anonymous

For the moment u can eliminate A and C

- anonymous

The problem with that is the function can never be U shaped, it can be a line and a V however

- anonymous

Can you change it?

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- anonymous

I THINK ITS B"

- marinos

The answer is (B) and (D).
The vertex of the graph of the absolute value function \[f(x)=a|x-h|+k\] is the point \[(h,k)\] and hence the he distance of the vertex from the x-axis is equal to \[|h|\] Neither of these depend on the value of \[a\], therefore we can eliminate choices (A) and (C). Now, depending on the sign of \[a\], the graph is V-shaped if \[a>0\], Λ-shaped if \[a>0\] and a straight (horizontal) line if \[a=0\] Interpreting Λ-shape as an "upside-down" V-shape, choice (B) is correct. (Note that the graph can never be U-shaped; however choice (B) is still valid, since we can have no values of the coefficient determining the U-shape, which is vacuously true.) Finally, the "width" of the graph in the case of the standard form of the) absolute value function makes sense only in comparison with the "width" of the graph of another absolute value function. We can normalize the notion of "width" with respect to the (simple) absolute value function \[y=|x|\] Thus, compared to the width of the latter, the graph of \[f(x)\] is wider if \[|a|>1\], narrower if \[|a|<1\] or has the same width if \[|a|=1\] Therefore, choice (D) is a correct answer.

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