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When graphed in the 
xy-plane, the function 
f(x)=3(2)^(x)+1 has a 
y-intercept at 
(0,k). What is the value of 
k ?
Choose 1 answer:
(A) 1
(B) 2
(c) 3
(D) 4

When graphed in the xy x y -plane, the function f(x)=3(2)x+1 f(x)=3(2)^{x}+1 has a y y -intercept at (0,k) (0, k) . What is the value of k k ?\newlineChoose 11 answer:\newline(A) 11\newline(B) 22\newline(C) 33\newline(D) 44

Full solution

Q. When graphed in the xy x y -plane, the function f(x)=3(2)x+1 f(x)=3(2)^{x}+1 has a y y -intercept at (0,k) (0, k) . What is the value of k k ?\newlineChoose 11 answer:\newline(A) 11\newline(B) 22\newline(C) 33\newline(D) 44
  1. Evaluate at x=0x=0: To find the y-intercept of the function, we need to evaluate the function at x=0x = 0. The y-intercept occurs where the graph of the function crosses the y-axis, which is when x=0x = 0.
  2. Substitute x=0x=0: Substitute x=0x = 0 into the function f(x)=3(2)x+1f(x) = 3(2)^x + 1 to find the value of kk.\newlinef(0)=3(2)0+1f(0) = 3(2)^0 + 1
  3. Calculate 202^0: Calculate the value of 22 raised to the power of 00, which is 11. \newline20=12^0 = 1
  4. Multiply by 33: Multiply 33 by the result from the previous step.\newline3×1=33 \times 1 = 3
  5. Add 11: Add 11 to the result from the previous step to find the value of f(0)f(0), which is the y-intercept kk.\newline3+1=43 + 1 = 4

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