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Functions - Properties of Polynomial Functions

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A polynomial function of degree nn is defined by the expression P(x)=anxn+anβˆ’1xnβˆ’1+β‹―+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0. Visually, these functions are always smooth and continuous curves with no breaks, holes, or sharp corners (cusps), extending infinitely in both directions along the x-axis.

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The end behavior of a polynomial is determined by its degree nn and leading coefficient ana_n. If nn is even, both ends of the graph point in the same vertical direction (forming a general 'U' or 'W' shape). If nn is odd, the ends point in opposite directions (forming a general 'S' or 'N' shape). Positive leading coefficients result in the right side of the graph pointing upwards.

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The yy-intercept of the polynomial is the point (0,a0)(0, a_0), where the graph crosses the vertical axis. This is found by evaluating P(0)P(0).

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Real zeros or roots are the xx-values where P(x)=0P(x) = 0, appearing as xx-intercepts on the graph. The multiplicity of a root kk affects the graph's behavior: if kk is odd, the graph crosses the xx-axis; if kk is even, the graph is tangent to the xx-axis (touches and turns around), creating a local maximum or minimum at that point.

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A polynomial of degree nn can have at most nn real roots and at most nβˆ’1n-1 turning points. Turning points are the 'peaks' (local maxima) and 'valleys' (local minima) where the function changes from increasing to decreasing or vice-versa.

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The Factor Theorem states that (xβˆ’c)(x - c) is a factor of P(x)P(x) if and only if P(c)=0P(c) = 0. Geometrically, this means the graph passes exactly through the point (c,0)(c, 0). The Remainder Theorem states that the remainder of P(x)xβˆ’c\frac{P(x)}{x-c} is equal to the value P(c)P(c).

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Vieta's Formulas relate the coefficients of a polynomial to sums and products of its roots. For a polynomial of degree nn, the sum of the roots is βˆ’anβˆ’1an-\frac{a_{n-1}}{a_n} and the product of the roots is (βˆ’1)na0an(-1)^n \frac{a_0}{a_n}.

πŸ“Formulae

P(x)=anxn+anβˆ’1xnβˆ’1+β‹―+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0

Remainder=P(c)Β whenΒ P(x)Β isΒ dividedΒ byΒ (xβˆ’c)\text{Remainder} = P(c) \text{ when } P(x) \text{ is divided by } (x-c)

(xβˆ’c)Β isΒ aΒ factorΒ β€…β€ŠβŸΊβ€…β€ŠP(c)=0(x-c) \text{ is a factor } \iff P(c) = 0

\text{Sum of roots } (\sum \alpha_i) = -\frac{a_{n-1}}{a_n}$

\text{Product of roots } (\prod \alpha_i) = (-1)^n \frac{a_0}{a_n}$

\text{Number of Turning Points} \le n - 1$

πŸ’‘Examples

Problem 1:

Find a cubic polynomial P(x)P(x) that has zeros at x=βˆ’1,2,Β andΒ 4x = -1, 2, \text{ and } 4, and passes through the point (1,12)(1, 12).

Solution:

  1. Using the zeros, write the factored form: P(x)=k(x+1)(xβˆ’2)(xβˆ’4)P(x) = k(x + 1)(x - 2)(x - 4). \ 2. Substitute the point (1,12)(1, 12) into the equation to find kk: \ 12=k(1+1)(1βˆ’2)(1βˆ’4)12 = k(1 + 1)(1 - 2)(1 - 4) \ 12=k(2)(βˆ’1)(βˆ’3)12 = k(2)(-1)(-3) \ 12=6kβ€…β€ŠβŸΉβ€…β€Šk=212 = 6k \implies k = 2. \ 3. Write the final equation: P(x)=2(x+1)(xβˆ’2)(xβˆ’4)P(x) = 2(x + 1)(x - 2)(x - 4). \ 4. (Optional) Expand to standard form: P(x)=2(x3βˆ’5x2+2x+8)=2x3βˆ’10x2+4x+16P(x) = 2(x^3 - 5x^2 + 2x + 8) = 2x^3 - 10x^2 + 4x + 16.

Explanation:

Since the zeros are given, the Factor Theorem allows us to construct the function using its linear factors. We include a leading coefficient kk and solve for it using the provided coordinate point.

Problem 2:

Factorize f(x)=x3βˆ’4x2+x+6f(x) = x^3 - 4x^2 + x + 6 completely given that x=2x = 2 is a zero.

Solution:

  1. Since x=2x = 2 is a zero, (xβˆ’2)(x - 2) is a factor. \ 2. Perform synthetic division or long division of (x3βˆ’4x2+x+6)(x^3 - 4x^2 + x + 6) by (xβˆ’2)(x - 2): \ Dividing gives the quadratic quotient x2βˆ’2xβˆ’3x^2 - 2x - 3. \ 3. Factorize the quadratic quotient: \ x2βˆ’2xβˆ’3=(xβˆ’3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1). \ 4. State the completely factorized form: f(x)=(xβˆ’2)(xβˆ’3)(x+1)f(x) = (x - 2)(x - 3)(x + 1). \ 5. The zeros are x=2,3,βˆ’1x = 2, 3, -1.

Explanation:

We use the Factor Theorem to identify the first linear factor. By dividing the cubic by this factor, we reduce the polynomial to a quadratic, which can then be solved using standard factoring techniques (finding numbers that multiply to βˆ’3-3 and add to βˆ’2-2).