High School Math 3

Try This First

Try this short activity before reading the lesson.

> Take $y=(x+1)(x-2)^2$. Before reading anything, plug in $x=-2$, $x=0$, $x=3$ and just write the sign of each answer. Then mark where $y=0$. You now have enough to draw the shape; do it, and keep the sketch.

Math lab: Predict the shape from signs

Predict first. For $f(x) = (x - 1)(x + 2)$, predict whether the graph is above or below the axis between its zeros.

<div class="math-model"><p><strong>Sign chart. Zeros at −2 and 1.</strong></p><div class="model-bar"><span class="model-cell">x &lt; −2: +</span><span class="model-cell">−2 &lt; x &lt; 1: −</span><span class="model-cell">x &gt; 1: +</span></div><p class="model-caption">Intervals are schematic, not drawn to scale.</p></div>

Model and solve. Test $x = -3, 0, 2$: outputs are $4, -2, 4$. The product is negative between the zeros. Plot the intercepts $(-2,0)$, $(1,0)$ and $(0,-2)$, then sketch the upward-opening parabola.

Spot the bug / explain. A learner says two factors always make a positive product. Use $x=0$ as a counterexample.

Your turn. Find zeros and the negative interval for $(x-2)(x+1)$.

Check after trying. Zeros: $-1$ and $2$; negative for $-1 < x < 2$.

The Big Idea

Far from the origin, only the leading term matters: at $x=1000$, the $x^4$ in $3x^4-5x^3+x$ is a thousand times the next term. So the degree and the sign of the leading coefficient decide the two ends. Even degree: both ends go the same way. Odd degree: opposite ways. Negative leading coefficient flips both.

Near the axis, the factors matter. A zero from a factor to an odd power crosses the axis; a zero from an even power touches and turns back, a bounce, because the sign cannot change there. Between zeros the sign is fixed, so one test point per interval settles it.

The graph can turn at most $n-1$ times for degree $n$. That is a ceiling, not a count.

The leading term decides the ends and the factors decide the middle.

Words to Know

• End behavior: Where the graph heads as $x\to\infty$ and as $x\to-\infty$. Decided by the leading term alone.
• Multiplicity: The power on a factor. $(x-1)^2$ gives the zero $1$ multiplicity $2$. Odd: cross. Even: bounce.
• Turning point: A place where the graph changes from rising to falling or back. Degree $n$ allows at most $n-1$.

Formulas and Theorems

• End behavior: $p(x)\sim a_n x^n \text{ for large } |x|$
• Note: Sign of $a_n$ and parity of $n$ decide where the ends point.
• Multiplicity: $(x-r)^m$
• Note: Crosses at $r$ if $m$ is odd, bounces if $m$ is even.
• Turning points: $\le n-1$
• Note: A degree-$n$ polynomial turns at most $n-1$ times.

Worked Examples and Your Turn

Study each example, then try the paired problem.

Example 1: The ends from the leading term

Problem: Describe the end behavior of $p(x)=3x^4-5x^3+x$.

Solution:

• Find the leading term: $3x^4$. Why: for large $|x|$ it swamps everything else, so it alone sets the ends.
• Degree $4$ is even, coefficient $3$ is positive. Why: $x^4$ is positive on both sides, and multiplying by $3$ keeps it positive.

Answer: Both ends go up: $p(x)\to\infty$ as $x\to\infty$ and as $x\to-\infty$.

> Verification & Check: $p(10)=30000-5000+10>0$ and $p(-10)=30000+5000-10>0$.

Your Turn 1: End behavior of $q(x)=-2x^3+7x$?

Example 2: Sketch from factored form

Problem: Sketch $p(x)=-(x+2)(x-1)^2(x-4)$ and state the sign of $p(0)$ without a calculator.

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fill="#333" text-anchor="start">(0, 8)</text><text x="181.3" y="33.2" font-family="Source Sans 3" font-size="12" fill="#1f5f8b" text-anchor="start">p(x) = −(x + 2)(x − 1)²(x − 4)</text></svg>

Solution:

• Degree $1+2+1=4$, leading coefficient $-1$. Why: multiply the leading $x$ of each factor; the minus in front makes both ends go down.
• Zeros: $-2$ (cross), $1$ (bounce, even power), $4$ (cross). Why: $(x-1)^2$ cannot change sign, so the graph touches at $1$ and turns.
• $p(0)=-(2)(1)(-4)=8$, positive. Why: one test point fixes the sign of the whole interval $(-2,1)$.
• Assemble: up from the far left through $-2$, above the axis, down to touch at $1$, back up, then down through $4$ and off to the bottom right. Why: the ends, the zeros and the sign at $0$ leave only one shape.

Answer: Both ends down; crosses at $-2$ and $4$, bounces at $1$ from above; $p(0)=8>0$. Three turning points, the most degree $4$ allows.

> Notice: Between $-2$ and $4$ the graph is never below the axis: $p(2)=-(4)(1)(-2)=8$ too. The bounce at $1$ only kisses it.

Your Turn 2: Sketch $q(x)=(x+1)(x-3)^2$. Ends, zeros, sign of $q(0)$?

Example 3: From the picture back to a formula

Problem: A cubic graph crosses the axis at $x=-3$, bounces at $x=2$, and passes through $(0,-24)$. Write $p(x)$.

The chapter continues in the book.