Chapter 5 Polynomials and Polynomial Functions - 5.7 Apply the Fundamental Theorem of Algebra - 5.7 Exercises - Mixed Review - Page 386: 78

$y=\dfrac{1}{3}(x-2)(x-7)$

Here, we find the x-intercepts are $2$ and $7$. Thus, the function is: $y=a (x-2)(x-7)$ Since $(4,-2)$ is on the graph of this function, we have $-2=a (x-2)(x-7)$ This gives: $-2=-6a \implies a=\dfrac{1}{3}$ Hence, the required function is: $y=\dfrac{1}{3}(x-2)(x-7)$