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

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

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

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