Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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)=((1)/(4))^(x)-1
(B) 
g(x)=2((2)/(3))x+1
(c) 
g(x)=2^(x)+3
(D) 
g(x)=3^(x)-2

The graph of y=g(x)+2 y=g(x)+2 in the xy x y -plane has a y y -intercept of 11 . If g g is an exponential function, which of the following could define g g ?\newlineChoose 11 answer:\newline(A) g(x)=(14)x1 g(x)=\left(\frac{1}{4}\right)^{x}-1 \newline(B) g(x)=2(23)x+1 g(x)=2\left(\frac{2}{3}\right) x+1 \newline(C) g(x)=2x+3 g(x)=2^{x}+3 \newline(D) g(x)=3x2 g(x)=3^{x}-2

Full solution

Q. The graph of y=g(x)+2 y=g(x)+2 in the xy x y -plane has a y y -intercept of 11 . If g g is an exponential function, which of the following could define g g ?\newlineChoose 11 answer:\newline(A) g(x)=(14)x1 g(x)=\left(\frac{1}{4}\right)^{x}-1 \newline(B) g(x)=2(23)x+1 g(x)=2\left(\frac{2}{3}\right) x+1 \newline(C) g(x)=2x+3 g(x)=2^{x}+3 \newline(D) g(x)=3x2 g(x)=3^{x}-2
  1. Set x=0x=0 and solve: To find the y-intercept of the function y=g(x)+2y=g(x)+2, we set x=0x=0 and solve for yy.\newliney=g(0)+2y = g(0) + 2\newlineWe are given that the y-intercept is 11, so:\newline1=g(0)+21 = g(0) + 2
  2. Find g(0)=1g(0) = -1: Solving for g(0)g(0), we get:\newlineg(0)=12g(0) = 1 - 2\newlineg(0)=1g(0) = -1\newlineSo, the function g(x)g(x) must satisfy the condition g(0)=1g(0) = -1.
  3. Check each option: Now we will check each option to see which one satisfies g(0)=1g(0) = -1.
    (A) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 1
    g(0)=(14)01g(0) = \left(\frac{1}{4}\right)^0 - 1
    g(0)=11g(0) = 1 - 1
    g(0)=0g(0) = 0
    This does not satisfy g(0)=1g(0) = -1.
  4. Check each option: Now we will check each option to see which one satisfies g(0)=1g(0) = -1.\newline(A) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 1\newlineg(0)=(14)01g(0) = \left(\frac{1}{4}\right)^0 - 1\newlineg(0)=11g(0) = 1 - 1\newlineg(0)=0g(0) = 0\newlineThis does not satisfy g(0)=1g(0) = -1.(B) g(x)=2(23)x+1g(x)=2\left(\frac{2}{3}\right)^x + 1\newlineg(0)=2(23)0+1g(0) = 2\left(\frac{2}{3}\right)^0 + 1\newlineg(0)=2(1)+1g(0) = 2(1) + 1\newlineg(0)=2+1g(0) = 2 + 1\newlineg(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 100\newlineThis does not satisfy g(0)=1g(0) = -1.
  5. Check each option: Now we will check each option to see which one satisfies g(0)=1g(0) = -1.
    (A) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 1
    g(0)=(14)01g(0) = \left(\frac{1}{4}\right)^0 - 1
    g(0)=11g(0) = 1 - 1
    g(0)=0g(0) = 0
    This does not satisfy g(0)=1g(0) = -1.
    (B) g(x)=2(23)x+1g(x)=2\left(\frac{2}{3}\right)^x + 1
    g(0)=2(23)0+1g(0) = 2\left(\frac{2}{3}\right)^0 + 1
    g(0)=2(1)+1g(0) = 2(1) + 1
    g(0)=2+1g(0) = 2 + 1
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 100
    This does not satisfy g(0)=1g(0) = -1.
    (C) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 122
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 133
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 144
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 155
    This does not satisfy g(0)=1g(0) = -1.
  6. Check each option: Now we will check each option to see which one satisfies g(0)=1g(0) = -1.
    (A) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 1
    g(0)=(14)01g(0) = \left(\frac{1}{4}\right)^0 - 1
    g(0)=11g(0) = 1 - 1
    g(0)=0g(0) = 0
    This does not satisfy g(0)=1g(0) = -1.
    (B) g(x)=2(23)x+1g(x)=2\left(\frac{2}{3}\right)^x + 1
    g(0)=2(23)0+1g(0) = 2\left(\frac{2}{3}\right)^0 + 1
    g(0)=2(1)+1g(0) = 2(1) + 1
    g(0)=2+1g(0) = 2 + 1
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 100
    This does not satisfy g(0)=1g(0) = -1.
    (C) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 122
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 133
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 144
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 155
    This does not satisfy g(0)=1g(0) = -1.
    (D) g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 177
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 188
    g(x)=(14)x1g(x)=\left(\frac{1}{4}\right)^x - 199
    g(0)=1g(0) = -1
    This satisfies g(0)=1g(0) = -1.

More problems from Compare linear and exponential growth