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The graph of y=g(x+5) in the xy-plane passes through the point (-5,4). If g is an exponential function, which of the following could define g ? Choose 1 answer: (A) g(x)=-5(4)^(x)+1 (B) g(x)=4^(x) (C) g(x)=4((4)/(5))^(x) (D) g(x)=4(2)^(x)-5

The graph of y=g(x+5)y=g(x+5) in the xyxy-plane passes through the point (−5,4)(-5,4). If gg is an exponential function, which of the following could define gg?\newlineChoose 11 answer:\newline(A) g(x)=−5(4)x+1g(x)=-5(4)^{x}+1\newline(B) g(x)=4xg(x)=4^{x}\newline(C) g(x)=4(45)xg(x)=4\left(\frac{4}{5}\right)^{x}\newline(D) g(x)=4(2)x−5g(x)=4(2)^{x}-5

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Q. The graph of y=g(x+5)y=g(x+5) in the xyxy-plane passes through the point (−5,4)(-5,4). If gg is an exponential function, which of the following could define gg?\newlineChoose 11 answer:\newline(A) g(x)=−5(4)x+1g(x)=-5(4)^{x}+1\newline(B) g(x)=4xg(x)=4^{x}\newline(C) g(x)=4(45)xg(x)=4\left(\frac{4}{5}\right)^{x}\newline(D) g(x)=4(2)x−5g(x)=4(2)^{x}-5
  1. Substitute x=−5x = -5: To find the correct function gg, we need to substitute x=−5x = -5 into the equation y=g(x+5)y = g(x + 5) and check which option gives y=4y = 4.
  2. Option (A): For option (A), g(x)=−5(4)x+1g(x) = -5(4)^x + 1, we substitute x=0x = 0 (since x+5=−5+5=0x + 5 = -5 + 5 = 0) to get g(0)=−5(4)0+1=−5(1)+1=−5+1=−4g(0) = -5(4)^0 + 1 = -5(1) + 1 = -5 + 1 = -4, which does not equal 44.
  3. Option (B): For option (B), g(x)=4xg(x) = 4^x, we substitute x=0x = 0 (since x+5=−5+5=0x + 5 = -5 + 5 = 0) to get g(0)=40=1g(0) = 4^0 = 1, which does not equal 44.
  4. Option (C): For option (C), g(x)=4(45)xg(x) = 4\left(\frac{4}{5}\right)^x, we substitute x=0x = 0 (since x+5=−5+5=0x + 5 = -5 + 5 = 0) to get g(0)=4(45)0=4(1)=4g(0) = 4\left(\frac{4}{5}\right)^0 = 4(1) = 4, which equals 44.
  5. Option (D): For option (D), g(x)=4(2)x−5g(x) = 4(2)^x - 5, we substitute x=0x = 0 (since x+5=−5+5=0x + 5 = -5 + 5 = 0) to get g(0)=4(2)0−5=4(1)−5=4−5=−1g(0) = 4(2)^0 - 5 = 4(1) - 5 = 4 - 5 = -1, which does not equal 44.

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