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Find g(x)g(x), where g(x)g(x) is the reflection across the yy-axis of f(x)=−4x+1f(x) = -4x + 1.\newlineChoices:\newline[g(x) = 4x - 1]\text{[g(x) = 4x - 1]}\newline[g(x) = -4x + 1]\text{[g(x) = -4x + 1]}\newline[g(x) = 4x + 1]\text{[g(x) = 4x + 1]}\newline[g(x) = -4x - 1]\text{[g(x) = -4x - 1]}

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Full solution

Q. Find g(x)g(x), where g(x)g(x) is the reflection across the yy-axis of f(x)=−4x+1f(x) = -4x + 1.\newlineChoices:\newline[g(x) = 4x - 1]\text{[g(x) = 4x - 1]}\newline[g(x) = -4x + 1]\text{[g(x) = -4x + 1]}\newline[g(x) = 4x + 1]\text{[g(x) = 4x + 1]}\newline[g(x) = -4x - 1]\text{[g(x) = -4x - 1]}
  1. Reflecting across y-axis: Reflecting a function across the y-axis means replacing xx with −x-x in the function's formula.f(x)=−4x+1f(x) = -4x + 1 becomes g(x)=−4(−x)+1g(x) = -4(-x) + 1 when reflected across the y-axis.
  2. Simplifying the expression: Simplify the expression for g(x)g(x).
    g(x)=−4(−x)+1g(x) = -4(-x) + 1 simplifies to g(x)=4x+1g(x) = 4x + 1.

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