## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{f(x)=5x^2-30x+40}$
RECALL: The zeros of the quadratic function $f(x) =c(x-a)(x-b)=0$ are $x=a$ and $x=b$. The given quadratic function has the zeros $x=4$ and $x=2$. Using the rule mentioned in the recall part above, then the function is: $f(x)=c(x-4)(x-2) \\f(x)=c[x(x)+x(-2) - 4(x) - 4(-2)] \\f(x)=c(x^2-2x-4x+8) \\f(x)=c(x^2-6x+8)$ With $f(0) = 40$, substitute $40$ to $f(0)$ and $0$ to $x$ to obtain: $f(0) = c(0^2-6\cdot 0+8) \\40 = c(0-0+8) \\40=c(8) \\\frac{40}{8}=c \\5=c$ Thus, the function is: $f(x) = 5(x^2-6x+8) \\\color{blue}{f(x)=5x^2-30x+40}$