Graphing from Factored Form

Consider $y = (x-1)(x-2)(x-3)$. Without drawing anything, decide whether the curve is above or below the horizontal axis when $x = 0$, when $x = 2.5$, and when $x = 10$. You only need the sign of each bracket, not its value.

A sketch needs three things and no more: the $x$-intercepts from the factors, the $y$-intercept from $P(0)$, and the end behaviour from the leading term alone.

To sketch $y = P(x)$ you need three things.

The $x$-intercepts. Set each factor to zero. For $y = (x+2)(x-1)(x-3)$ the curve meets the horizontal axis at $-2$, $1$ and $3$.

The $y$-intercept. Substitute $x = 0$. Here $P(0) = (2)(-1)(-3) = 6$, so the curve passes through $(0,6)$. In expanded form this is just the constant term.

The end behaviour. Multiply out only the leading terms: $x \times x \times x = x^3$. Degree $3$ is odd and the coefficient is positive, so the curve comes up from below on the left and continues upward on the right.

That is enough for a sketch. Everything else, the exact positions of the turning points and the exact steepness, is beyond what Year 10 methods can find and beyond what a sketch is asked to show.

Take $P(x) = x^3 - 100x$. Near the origin the $-100x$ term is much the larger of the two. But compare the two terms at $x = 1000$:

$$x^3 = 1\,000\,000\,000, \qquad 100x = 100\,000$$

The cubic term is ten thousand times bigger, and the gap widens without limit as $x$ grows, because each extra power of $x$ multiplies the leading term by another factor of $x$ while the lower terms gain nothing comparable.

So far from the origin, a polynomial behaves like its leading term and nothing else. Two features of that term settle the ends:

The parity of the degree. Even powers of $x$ are positive at both ends; odd powers change sign. So an even degree sends both ends the same way, and an odd degree sends them opposite ways.

The sign of the leading coefficient. A negative coefficient flips both ends.

Those two facts give the four cases in the diagram above, and no other information about the polynomial can change which case applies.

Between two consecutive zeroes, a polynomial cannot change sign. Changing from positive to negative would require passing through zero, and there is no zero in between.

So the zeroes cut the number line into intervals, and the sign is constant on each. Testing one convenient point per interval settles it.

For $y = (x+2)(x-1)(x-3)$ the zeroes $-2$, $1$, $3$ give four intervals. Take one test point in each and record only the signs of the brackets:

$x = -3$: $(-)(-)(-) = -$, so the curve is below the axis.
$x = 0$: $(+)(-)(-) = +$, above.
$x = 2$: $(+)(+)(-) = -$, below.
$x = 4$: $(+)(+)(+) = +$, above.

The curve therefore comes up from below on the left, rises through $-2$, turns over and comes back down through $1$, turns again and rises through $3$, and continues up. Which is exactly what the end behaviour predicted, so the two methods have checked each other.

Never compute the values, only the signs. $(+)(-)(-)$ is positive whatever the sizes are, and stopping at the sign is several times faster.

A polynomial given in expanded form has to be factorised before any of this works. Sketch $y = x^3 - 3x^2 - 4x$.

Factorise. Take out the common factor, then handle the quadratic:

$$y = x(x^2 - 3x - 4) = x(x-4)(x+1)$$

Intercepts. $x$-intercepts at $-1$, $0$ and $4$. The $y$-intercept is $P(0) = 0$, so the curve passes through the origin, which the factor of $x$ had already told us.

Ends. The leading term is $x^3$: odd degree, positive coefficient, so down on the left and up on the right.

Signs. Testing $x = -2$: $(-)(-)(-) = -$. Testing $x = -0.5$: $(-)(-)(+) = +$. Testing $x = 1$: $(+)(-)(+) = -$. Testing $x = 5$: $(+)(+)(+) = +$.

The sketch now draws itself: below, up through $-1$, above, down through $0$, below, up through $4$, and away.

A quartic works identically, with one more zero and one more interval. If a cubic factor remains after the first division, factorise that too; the method never changes.

Two honest limitations are worth stating, because pretending otherwise leads to wrong answers.

Turning points. The methods here locate the crossings, not the peaks and valleys between them. For $y = x(x-4)(x+1)$ the turning point between $0$ and $4$ is not at $x = 2$; finding it exactly needs calculus, which is a Year 11 topic. On a sketch it goes roughly in the middle, and that is accepted.

Vertical scale. A cubic's values grow fast, so a sketch that is honest about the horizontal positions is usually dishonest about the heights. That is normal, and the marking allows it, provided the intercepts and the order are right.

What a sketch must get right: every $x$-intercept, in the correct order; the $y$-intercept; the correct side of the axis on every interval; and both ends pointing the correct way. Those four are checkable and are where the marks are.

Five items on sketching. Work each one, then reveal the answer.

1. 1 State the $x$-intercepts of $y = (x-5)(x+2)(2x-1)$.

   Show answer
   Set each bracket to zero and solve.$5$, $-2$ and $\tfrac{1}{2}$

2. 2 State the $y$-intercept of $y = (x+1)(x-4)(x+3)$.

   Show answer
   Substitute $x = 0$: $(1)(-4)(3)$.$-12$

3. 3 Describe the ends of $y = -2x^4 + x - 7$.

   Show answer
   Even degree, negative leading coefficient.Both ends down

4. 4 Describe the ends of $y = 5x^7 - x^2$.

   Show answer
   Odd degree, positive leading coefficient.Down-left, up-right

5. 5 Is $y = (x-1)(x-2)(x-3)$ above or below the axis at $x = 2.5$?

   Show answer
   Signs only: $(+)(+)(-)$.Below