If a quadratic function in $x$ has the value 19 when $x=1$ and has a maximum value 20 when $x=2$, then the function is

Let the quadratic function be $f(x) = ax^2 + bx + c$.

Since the function has a maximum value at $x=2$, the coefficient $a$ must be negative, and the vertex is at $x = -\dfrac{b}{2a} = 2$, which implies $b = -4a$.

The maximum value is $f(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 20$.

Substituting $b = -4a$ into the equation for the maximum value: $4a + 2(-4a) + c = 20 \Rightarrow -4a + c = 20 \Rightarrow c = 20 + 4a$.

Given $f(1) = 19$, we have $a(1)^2 + b(1) + c = 19 \Rightarrow a + b + c = 19$.

Substituting $b = -4a$ and $c = 20 + 4a$ into this equation: $a - 4a + 20 + 4a = 19 \Rightarrow a + 20 = 19 \Rightarrow a = -1$.

Now find $b$ and $c$: $b = -4(-1) = 4$ and $c = 20 + 4(-1) = 16$.

The function is $f(x) = -x^2 + 4x + 16$.

Answer: $f(x)=-x^2+4 x+16$