The graph of y=g(x)+2 in the xy-plane has a y-intercept of 1 . If g is an exponential function, which of the following could define g ?Choose 1 answer:(A) g(x)=(41)x−1(B) g(x)=2(32)x+1(C) g(x)=2x+3(D) g(x)=3x−2
Q. The graph of y=g(x)+2 in the xy-plane has a y-intercept of 1 . If g is an exponential function, which of the following could define g ?Choose 1 answer:(A) g(x)=(41)x−1(B) g(x)=2(32)x+1(C) g(x)=2x+3(D) g(x)=3x−2
Set x=0 and solve: To find the y-intercept of the function y=g(x)+2, we set x=0 and solve for y.y=g(0)+2We are given that the y-intercept is 1, so:1=g(0)+2
Find g(0)=−1: Solving for g(0), we get:g(0)=1−2g(0)=−1So, the function g(x) must satisfy the condition g(0)=−1.
Check each option: Now we will check each option to see which one satisfies g(0)=−1. (A) g(x)=(41)x−1 g(0)=(41)0−1 g(0)=1−1 g(0)=0 This does not satisfy g(0)=−1.
Check each option: Now we will check each option to see which one satisfies g(0)=−1.(A) g(x)=(41)x−1g(0)=(41)0−1g(0)=1−1g(0)=0This does not satisfy g(0)=−1.(B) g(x)=2(32)x+1g(0)=2(32)0+1g(0)=2(1)+1g(0)=2+1g(x)=(41)x−10This does not satisfy g(0)=−1.
Check each option: Now we will check each option to see which one satisfies g(0)=−1. (A) g(x)=(41)x−1 g(0)=(41)0−1 g(0)=1−1 g(0)=0 This does not satisfy g(0)=−1. (B) g(x)=2(32)x+1 g(0)=2(32)0+1 g(0)=2(1)+1 g(0)=2+1 g(x)=(41)x−10 This does not satisfy g(0)=−1. (C) g(x)=(41)x−12 g(x)=(41)x−13 g(x)=(41)x−14 g(x)=(41)x−15 This does not satisfy g(0)=−1.
Check each option: Now we will check each option to see which one satisfies g(0)=−1. (A) g(x)=(41)x−1 g(0)=(41)0−1 g(0)=1−1 g(0)=0 This does not satisfy g(0)=−1. (B) g(x)=2(32)x+1 g(0)=2(32)0+1 g(0)=2(1)+1 g(0)=2+1 g(x)=(41)x−10 This does not satisfy g(0)=−1. (C) g(x)=(41)x−12 g(x)=(41)x−13 g(x)=(41)x−14 g(x)=(41)x−15 This does not satisfy g(0)=−1. (D) g(x)=(41)x−17 g(x)=(41)x−18 g(x)=(41)x−19 g(0)=−1 This satisfies g(0)=−1.
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