1. The problem is to find a function $f(x)$ such that $f(0)=7$, $f(1)=4$, and $f(2)=3$. 2. We can try to find a polynomial function that fits these points. Since we have three points, a quadratic polynomial $f(x)=ax^2+bx+c$ is a good candidate. 3. Using the points, we set up the system: $$f(0)=c=7$$ $$f(1)=a+b+c=4$$ $$f(2)=4a+2b+c=3$$ 4. Substitute $c=7$ into the other equations: $$a+b+7=4 \implies a+b=-3$$ $$4a+2b+7=3 \implies 4a+2b=-4$$ 5. Solve the system: From $a+b=-3$, we get $b=-3 - a$. Substitute into $4a+2b=-4$: $$4a + 2(-3 - a) = -4$$ $$4a - 6 - 2a = -4$$ $$2a - 6 = -4$$ $$2a = 2 \implies a=1$$ 6. Then $b = -3 - 1 = -4$. 7. The quadratic function is: $$f(x) = 1 \cdot x^2 - 4x + 7 = x^2 - 4x + 7$$ 8. Verify: $$f(0) = 0 - 0 + 7 = 7$$ $$f(1) = 1 - 4 + 7 = 4$$ $$f(2) = 4 - 8 + 7 = 3$$ All points match. Final answer: $$f(x) = x^2 - 4x + 7$$