Invite students to share their responses and reasoning. If not mentioned in students’ explanations, emphasize that:

• $b$ and $d$ must be 0 because the $y$-coordinate of the $x$-intercepts of the graph of any function is 0.
• $e$ must be 0 because the $x$-coordinate of the $y$-intercept of the graph of any function is 0.
• $a$ and $c$ correspond to the 1.6 and 2 in the factored expression. Students may reason that the graph shows that the positive $x$-intercept is farther away from 0 than the negative $x$-intercept, suggesting that $c$ is 2 and $a$ is -1.6. They may also use their observations from an earlier lesson: that when a factor in the expression shows a negative sign or subtraction, a corresponding intercept takes a positive value, and when a factor shows a positive sign or addition, a corresponding intercept takes a negative value. Either explanation is reasonable at this point.
• $(0,f)$ is the $y$-intercept, so the value of $f$ can be found by evaluating $w(0)$, which is -3.2, because $(0+1.6)(0-2) = (1.6)(\text-2) = \text-3.2$.

Some students may think that the numerical values in the equation correspond directly to the $x$-intercepts in the graph and incorrectly state that $a= \text-2$ and $c=1.6$. Remind them that a graph shows all pairs of $x$ and $y$ values that make the equation true. Consider asking these students to try substituting -2 for $x$ and evaluating the expression to verify that $w(\text-2)=0$.

Students will have opportunities to attend to the signs or the operations in quadratic expressions in factored form, so it is not essential that this misconception is corrected at this moment.