A function f(t) increases by 4 over every unit interval in t and f(0) = 0. Which could be a function rule for f(t)? Choices: (A)f(t) = 4t (B)f(t) = -t - 4 (C)f(t) = -t + 4 (D)f(t) = -t/4

Plug t = 0: Since f(0) = 0, let's plug t = 0 into each choice to see which one gives us f(0) = 0. (A) f(0) = 4*0 = 0 (B) f(0) = -0 - 4 = -4 (C) f(0) = -0 + 4 = 4 (D) f(0) = -0/4 = 0 Choices (B) and (C) are out cuz they don't give us f(0) = 0.Check f(0) = 0: Now, we need to check which function increases by 4 for each unit increase in t. Let's check the remaining choices by plugging in t = 1. (A) f(1) = 4*1 = 4 (D) f(1) = -1/4 Only choice (A) increases by 4 when t increases by 1.

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

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Q. A function

f(t)

increases by

4

over every unit interval in

t

and

f(0) = 0

\newline

Which could be a function rule for

f(t)

\newline

Choices:

\newline

(A)

f(t) = 4t

\newline

(B)

f(t) = -t - 4

\newline

(C)

f(t) = -t + 4

\newline

(D)

f(t) = -\frac{t}{4}

Plug $t = 0$ : Since $f(0) = 0$ , let's plug $t = 0$ into each choice to see which one gives us $f(0) = 0$ .
(A) $f(0) = 4\times0 = 0$
(B) $f(0) = -0 - 4 = -4$
(C) $f(0) = -0 + 4 = 4$
(D) $f(0) = -0/4 = 0$
Choices (B) and (C) are out cuz they don't give us $f(0) = 0$ .
Check $f(0) = 0$ : Now, we need to check which function increases by $4$ for each unit increase in $t$ . Let's check the remaining choices by plugging in $t = 1$ . $\newline$ (A) $f(1) = 4 \times 1 = 4$ $\newline$ (D) $f(1) = -\frac{1}{4}$ $\newline$ Only choice (A) increases by $4$ when $t$ increases by $1$ .