Topic 1.3 · Unit 1 · CA CCSS F-IF.7c, A-APR.3
Graphing polynomial functions
You are not plotting points. A polynomial graph is determined by three things (where it crosses, what it does at each crossing, and where the ends go) and a sketch that gets those right earns the marks.
- Factor and read the zeros. Each factor \( (x - c) \) puts an \( x \)-intercept at \( c \).
- Multiplicity decides the behavior there. Odd multiplicity crosses the axis; even multiplicity touches and turns back. Multiplicity 3 or more flattens as it crosses.
- End behavior comes from the leading term only. Even degree sends both ends the same way, odd degree sends them opposite ways; a positive leading coefficient sends the right end up, a negative one sends it down.
- The \( y \)-intercept is \( f(0) \), the constant term.
- Turning points: a degree-\( n \) polynomial has at most \( n - 1 \) of them. Use that as a check on your sketch.
- Sign chart: the sign can only change at a zero, so test one value in each interval to know whether the curve is above or below the axis there.
Where marks are lost: drawing every intercept as a crossing. In \( y = (x-1)^2(x+3) \), the graph touches at \( x = 1 \) and turns around; it does not pass through.
Example 1: a full sketch. Describe the graph of \( f(x) = (x+2)(x-1)^2 \).
Zeros at \( x = -2 \) (multiplicity 1, crosses) and \( x = 1 \) (multiplicity 2, touches).
Expanding mentally, the leading term is \( x^3 \): odd degree, positive coefficient, so the left end goes down and the right end goes up.
\( y \)-intercept: \( f(0) = (2)(1) = 2 \). The curve rises from the lower left, crosses at \( -2 \), passes through \( (0, 2) \), comes down to touch the axis at \( x = 1 \), and rises again.
Example 2: end behavior. Describe the end behavior of \( f(x) = -2x^4 + 7x^3 - x + 5 \).
Only \( -2x^4 \) matters. Even degree means both ends behave alike; the negative coefficient sends them down.
As \( x \to \pm\infty \), \( f(x) \to -\infty \). The other terms change the middle of the graph, never the ends.
Example 3: sign chart. Where is \( f(x) = x(x-3)(x+1) \) positive?
Zeros at \( -1, 0, 3 \) split the line into four intervals. Test one point in each: \( f(-2) = (-2)(-5)(-1) = -10 \) negative; \( f(-0.5) = (-0.5)(-3.5)(0.5) \) positive; \( f(1) = (1)(-2)(2) = -4 \) negative; \( f(4) = (4)(1)(5) = 20 \) positive.
So \( f(x) > 0 \) on \( (-1, 0) \) and \( (3, \infty) \).
1–5 zeros and end behavior, 6–11 full descriptions, 12–15 reading graphs and diagnosis.
- State the zeros of \( f(x) = (x-4)(x+2)(x-1) \).
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\( x = 4, -2, 1 \)
- State the degree of \( f(x) = (x-1)^2(x+3)^3 \).
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Add the multiplicities: \( 2 + 3 \). 5
- Describe the end behavior of \( f(x) = 3x^5 - x^2 + 1 \).
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Odd degree, positive leading coefficient. Down on the left, up on the right
- Describe the end behavior of \( f(x) = -x^6 + 4x \).
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Even degree, negative coefficient. Both ends down
- Find the \( y \)-intercept of \( f(x) = (x-2)(x+5) \).
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\( f(0) = (-2)(5) \). \( -10 \)
- At \( x = -3 \), does \( f(x) = (x+3)^2(x-1) \) cross or touch?
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Multiplicity 2, which is even. Touches and turns
- At \( x = 2 \), does \( f(x) = (x-2)^3(x+1) \) cross or touch?
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Multiplicity 3 is odd, so it crosses, flattening as it does. Crosses
- What is the maximum number of turning points of a degree-6 polynomial?
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\( n - 1 \). 5
- Fully describe the graph of \( f(x) = (x-1)(x+2)^2 \).
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Degree 3, positive leading coefficient, so down-left and up-right. Crosses at \( x = 1 \), touches at \( x = -2 \), \( y \)-intercept \( (-1)(4) = -4 \). See description
- Fully describe the graph of \( f(x) = -x^2(x-3) \).
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Leading term \( -x^3 \): up-left, down-right. Touches at \( x = 0 \), crosses at \( x = 3 \), \( y \)-intercept 0. See description
- Where is \( f(x) = (x-2)(x+1) \) negative?
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An upward parabola is below the axis between its roots. \( -1 < x < 2 \)
- Solve \( x(x-4)(x+2) > 0 \).
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Zeros at \( -2, 0, 4 \). Testing: \( x=-3 \) negative, \( x=-1 \) positive, \( x=1 \) negative, \( x=5 \) positive. \( -2 < x < 0 \) or \( x > 4 \)
- A graph touches the axis at \( x = 2 \), crosses at \( x = -1 \) and at \( x = 5 \), and
both ends go up. Write a possible equation.
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Touching needs even multiplicity, so \( (x-2)^2 \). Each crossing needs odd multiplicity, so \( (x+1) \) and \( (x-5) \). That is degree 4, even, and both ends go up with a positive leading coefficient, which matches. \( f(x) = (x-2)^2(x+1)(x-5) \)
- A student says \( f(x) = -x^4 + 100x^3 \) goes up on the right because of the
\( 100x^3 \). Find the error.
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End behavior depends only on the highest-degree term. For large \( x \), \( x^4 \) dominates \( 100x^3 \) no matter how big the coefficient, the graph turns and both ends go down. Both ends down
- A student draws \( y = (x-1)^2(x+3) \) crossing the axis at \( x = 1 \). Find the error.
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The factor is squared, so the multiplicity is even and the curve touches at \( x = 1 \) and turns back. Only \( x = -3 \) is a crossing. It touches, not crosses