We know that if $f(x)=x^2-bx+c$ then $f(a)=f(b-a)$. Thus if the consecutive integers $x-1, x, x+1$ are not $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$ then one of the three consecutive numbers is at least a distance of $\frac{3}{2}$ from $\frac{b}{2}$. Let this number be $y$. Then $f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$. Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. So we have another integer input to the function with a prime output. So now we assume that the 3 numbers are in fact $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$. These are integers only if $b=2k$ or $f(x)=x^2-2kx+c=(x-k)^2+(c-k^2)$. We know that at $k-1, k, k+1$ these are primes. But $f(k+1)-f(k)=((k+1)-k)^2+(c-k^2)-(k-k)^2-(c-k^2)=1$. The only two primes that are 1 apart are $2$ and $3$. Thus $f(k)=2$, so $c-k^2=2$ and $f(x)=(x-k)^2+2$. Then $f(k+3)=(k+3-k)^2+2=11$ which is another prime value at an integer point.

tenniskidperson3 wrote: ... Let this number be $y$. Then $f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$. Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. ... Why must $b-y$ be one of the 3 consecutive integers?

There is no reason why it should be; in fact it isn't. I don't understand what you're saying; $b-y$ is at least 3 away from $y$ which is one of the consecutive numbers, so $b-y$ cannot be.

Suppose $f(n-1),f(n),f(n+1)$ are primes. Define another polynomial $g(x)=f(x+n)$. So $g$ is a monic quadratic polynomial with integer coefficients such that $g(-1),g(0),g(1)$ are primes. It is enough to show that $g(x)$ attains a prime value at another integer point. Let $g(x)=x^2+ax+b$. Note that $g(x)=g(-a-x)$. So $g(-a-1),g(-a),g(-a+1)$ are primes. If the set $\{-a-1,-a,-a+1\}$ does not coincide with $\{-1,0,1\}$, then we're done. Otherwise, we get $a=0$. Then $g(-1)=1+b$ and $g(0)=b$, both of which are primes. So $b=2$ and $g(x)=x^2+2$, which attains a prime value at $g(3)=11$.