In the linear function h , h ( 0 ) = 41 and h ( 1 ) = 40 .

Question

In the linear function h , h ( 0 ) = 41 and h ( 1 ) = 40 . Which equation defines h ?

Accepted Answer

Choice A is correct. An equation defining a linear function can be written in the form $h (x) = a x + b$ , where $a$ and $b$ are constants. It’s given that $h left parenthesis 0 right parenthesis equals 41$ . Substituting $0$ for $x$ and $41$ for $h (x)$ in the equation $h (x) = a x + b$ yields $41 = a (0) + b$ , or $b equals 41$ . Substituting $41$ for $b$ in the equation $h (x) = a x + b$ yields $h (x) = a x + 41$ . It’s also given that $h left parenthesis 1 right parenthesis equals 40$ . Substituting $1$ for $x$ and $40$ for $h (x)$ in the equation $h (x) = a x + 41$ yields $40 = a (1) + 41$ , or $40 = a + 41$ . Subtracting $41$ from the left- and right-hand sides of this equation yields $- 1 = a$ . Substituting $negative 1$ for $a$ in the equation $h (x) = a x + 41$ yields $h (x) = - 1 x + 41$ , or $h (x) = - x + 41$ .

Choice B is incorrect. Substituting $0$ for $x$ and $41$ for $h (x)$ in this equation yields $41 = - 0$ , which isn't a true statement.

Choice C is incorrect. Substituting $0$ for $x$ and $41$ for $h (x)$ in this equation yields $41 = - 41 (0)$ , or $41 equals 0$ , which isn't a true statement.

Choice D is incorrect. Substituting $41$ for $h (x)$ in this equation yields $41 = - 41$ , which isn't a true statement.

Question in-the

In the linear function h , h ( 0 ) = 41 and h ( 1

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