Matching a polynomial graph to its factored form

Read three things off the graph.

1. Zeros. The graph meets the $x$-axis at one negative $x$-value well to the left of the origin, at one small positive value just right of the origin, and at a larger positive value.
2. Multiplicity. At the larger positive zero the graph touches the axis and turns back up without crossing. That means an even power on that factor, at least squared. At the other two zeros it crosses, which fits a single factor.
3. End behavior. Both ends go up, so the degree is even with a positive leading coefficient.

Choice C, $0.25(x-5{)}^2(x-1)(x+8)$, has zeros at $-8$ (crossing), $1$ (crossing) and $5$ (touching), degree $2+1+1=4$, which is even, and leading coefficient $0.25>0$. All three match.

As a check, its $y$-intercept is $0.25(-5{)}^2(-1)(8)=0.25\cdot 25\cdot (-8)=-50$, which is negative, and the graph does cross the $y$-axis below the origin.

Each choice

• A. The zeros are in the right places, but every factor is to the first power. That gives a cubic, which crosses the axis at all three zeros and has ends going in opposite directions. The graph touches at the right-hand zero and rises on both sides.
• B. Each factor has the wrong sign inside, so the zeros land at $-5$, $-1$ and $8$. This comes from reading $(x+5)$ as having a zero at 5. The graph has two positive zeros and one negative one, not the other way round.
• C (correct). A squared factor at $x=5$ gives the touch, single factors at 1 and $-8$ give the crossings, and an even degree with a positive coefficient sends both ends up.
• D. The repeated factor and even degree are right, but every sign inside the parentheses is flipped, putting the zeros at $-5$, $-1$ and $8$. That would place the touching zero on the negative side.