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Jacy says that f(x) = 4|x – 1| and f(x) = |4x – 1| have the same graph. Is Jacy correct? Explain

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4 votes

Jacy's statement that f(x) = 4|x – 1| and f(x) = |4x – 1| have the same graph is false

Hwo to determine the true statement

From the question, we have the following parameters that can be used in our computation:

f(x) = 4|x – 1| and f(x) = |4x – 1|

Opening the bracket of f(x) = 4|x – 1|, we have

f(x) = |4x – 4|

By comparison, we have that

f(x) = |4x – 4| and f(x) = |4x – 1| are not the same

Hene, Jacy's statement about the functions is false

User Bowen Xu
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