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A function f(t)f(t) increases by 44 over every unit interval in tt and f(0)=0f(0) = 0. Which could be a function rule for f(t)f(t)?\newlineChoices:\newline(A) f(t)=4tf(t) = 4t\newline(B) f(t)=t4f(t) = -t - 4\newline(C) f(t)=t+4f(t) = -t + 4\newline(D) f(t)=t4f(t) = -\frac{t}{4}

Full solution

Q. A function f(t)f(t) increases by 44 over every unit interval in tt and f(0)=0f(0) = 0. Which could be a function rule for f(t)f(t)?\newlineChoices:\newline(A) f(t)=4tf(t) = 4t\newline(B) f(t)=t4f(t) = -t - 4\newline(C) f(t)=t+4f(t) = -t + 4\newline(D) f(t)=t4f(t) = -\frac{t}{4}
  1. Plug t=0t = 0: Since f(0)=0f(0) = 0, let's plug t=0t = 0 into each choice to see which one gives us f(0)=0f(0) = 0.
    (A) f(0)=4×0=0f(0) = 4\times0 = 0
    (B) f(0)=04=4f(0) = -0 - 4 = -4
    (C) f(0)=0+4=4f(0) = -0 + 4 = 4
    (D) f(0)=0/4=0f(0) = -0/4 = 0
    Choices (B) and (C) are out cuz they don't give us f(0)=0f(0) = 0.
  2. Check f(0)=0f(0) = 0: Now, we need to check which function increases by 44 for each unit increase in tt. Let's check the remaining choices by plugging in t=1t = 1.\newline(A) f(1)=4×1=4f(1) = 4 \times 1 = 4\newline(D) f(1)=14f(1) = -\frac{1}{4}\newlineOnly choice (A) increases by 44 when tt increases by 11.

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