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Solved Examples on Polynomial Graphs

Calculators: Calculators for Polynomials

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(1.) (a.) Determine whether the function is a polynomial.
(b.) If it is a polynomial, write it in standard form.
(c.) State the degree.
(d.) Identify the leading term
(e.) Identify the constant term. $$ G(x) = 2(x - 1)^2(x^2 + 5) $$

(a.) The function is a polynomial

$ (b.) \\[3ex] G(x) = 2(x - 1)^2(x^2 + 5) \\[3ex] (x - 1)^2 \\[3ex] = (x - 1)(x - 1) \\[3ex] = x^2 - x - x + 1 \\[3ex] = x^2 - 2x + 1 \\[3ex] 2(x - 1)^2 \\[3ex] = 2(x^2 - 2x + 1) \\[3ex] = 2x^2 - 4x + 2 \\[3ex] 2(x - 1)^2(x^2 + 5) \\[3ex] = (2x^2 - 4x + 2)(x^2 + 5) \\[3ex] = 2x^4 + 10x^2 - 4x^3 - 20x + 2x^2 + 10 \\[3ex] = 2x^4 - 4x^3 + 12x^2 - 20x + 10 \\[3ex] \implies \\[3ex] G(x) = 2x^4 - 4x^3 + 12x^2 - 20x + 10 \\[3ex] $ (c.) The degree is: 4

(d.) The leading term is: 2x$^4$

(e.) The constant term is: 10

(2.) Determine the polynomial whose:
(a.) zeros are: −4, 4, 8 and degree is: 3
(b.) zeros are: −2, 0, 3 and degree is: 3
(c.) zeros are: −9, multiplicity 1; −4, multiplicity 2 and degree is: 3

$ (a.) \\[3ex] degree = 3 \\[3ex] zeros = -4, 4, 8 \\[3ex] factors = (x + 4)(x - 4)(x - 8) \\[3ex] \implies \\[3ex] f(x) = (x + 4)(x - 4)(x - 8) \\[3ex] (x + 4)(x - 4) = x^2 - 4^2 ...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \implies \\[3ex] f(x) = (x^2 - 4^2)(x - 8) \\[3ex] = (x^2 - 16)(x - 8) \\[3ex] = x^3 - 8x^2 - 16x + 128 (b.) \\[3ex] degree = 3 \\[3ex] zeros = -2, 0, 3 \\[3ex] factors = (x + 2)(x - 0)(x - 3) \\[3ex] factors = x(x + 2)(x - 3) \\[3ex] \implies \\[3ex] f(x) = x(x + 2)(x - 3) \\[3ex] = (x^2 + 2x)(x - 3) \\[3ex] = x^3 - 3x^2 + 2x^2 - 6x \\[3ex] = x^3 - x^2 - 6x \\[3ex] (c.) \\[3ex] degree = 3 \\[3ex] zeros = -9, -4, -4 \\[3ex] factors = (x + 9)(x + 4)(x + 4) \\[3ex] \implies \\[3ex] f(x) = (x + 9)(x + 4)(x + 4) \\[3ex] = (x + 9)(x^2 + 4x + 4x + 16) \\[3ex] = (x + 9)(x^2 + 8x + 16) \\[3ex] = x^3 + 8x^2 + 16x + 9x^2 + 72x + 144 \\[3ex] = x^3 + 17x^2 + 88x + 144 $

(3.) Determine the polynomial whose:
(a.) zeros are: −5, −1, 1, 2 and degree is: 4

$ (a.) \\[3ex] degree = 4 \\[3ex] zeros = -5, -1, 1, 2 \\[3ex] factors = (x + 5)(x + 1)(x - 1)(x - 2) \\[3ex] factors = (x + 1)(x - 1)(x + 5)(x - 2) \\[3ex] \implies \\[3ex] f(x) = (x + 1)(x - 1)(x + 5)(x - 2) \\[3ex] (x + 1)(x - 1) = x^2 - 1^2 ...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \implies \\[3ex] f(x) = (x^2 - 1^2)(x^2 - 2x + 5x - 10) \\[3ex] = (x^2 - 1)(x^2 + 3x - 10) \\[3ex] = x^4 + 3x^3 - 10x^2 - x^2 - 3x + 10 \\[3ex] = x^4 + 3x^3 - 11x^2 - 3x + 10 $

(4.) For the polynomial: G(x) = (x − 1)³(x + 1)
(a.) Determine the x-intercepts of the graph of G.
(b.) Determine the x-intercepts of the graph of G(x + 4)

To find x-intercept; set G(x) = 0 and solve for x

$ G(x) = (x - 1)^3(x + 1) \\[3ex] (a.) \\[3ex] (x - 1)^3(x + 1) = 0 \\[3ex] (x - 1)^3 = 0 \;\;\;OR\;\;\; x + 1 = 0...Zero\;\;Product\;\;Property \\[3ex] x - 1 = \sqrt[3]{0} \;\;\;OR\;\;\; x = -1 \\[3ex] x - 1 = 0 \;\;\;OR\;\;\; x = -1 \\[3ex] x = 1 \;\;\;OR\;\;\; x = -1 \\[3ex] x-intercepts = (-1, 0)\;\;and\;\;(1, 0) \\[3ex] (b.) \\[3ex] G(x + 4) = (x + 4 - 1)^3(x + 4 + 1) \\[3ex] = (x + 3)^3(x + 5) \\[3ex] \implies \\[3ex] (x + 3)^3 = 0 \;\;\;OR\;\;\; x + 5 = 0...Zero\;\;Product\;\;Property \\[3ex] x + 3 = \sqrt[3]{0} \;\;\;OR\;\;\; x = -5 \\[3ex] x + 3 = 0 \;\;\;OR\;\;\; x = -5 \\[3ex] x = -3 \;\;\;OR\;\;\; x = -5 $

(5.) Determine the polynomial function whose
(a.) zeros are: −3, 2, 5 and whose graph passes through the point (6, 72)

(b.) zeros are: −2, multiplicity 2; 2, multiplicity 2; and degree is: 4 and whose graph passes through the point (−3, 125)

$ (a.) \\[3ex] zeros = -3, 2, 5 \\[3ex] factors = (x + 3)(x - 2)(x - 5) \\[3ex] \implies \\[3ex] f(x) = a(x + 3)(x - 2)(x - 5) \\[3ex] passes\;\;through\;\;(6, 72) \\[3ex] To\;\;find\;\;leading\;\;coefficient:a \\[3ex] x = 6,\;\;\;f(x) = 72 \\[3ex] \implies \\[3ex] 72 = a(6 + 3)(6 - 2)(6 - 5) \\[3ex] 72 = a(9)(4)(1) \\[3ex] 36a = 72 \\[3ex] a = \dfrac{72}{36} \\[5ex] a = 2 \\[3ex] \implies \\[3ex] f(x) = a(x + 3)(x - 2)(x - 5) \\[3ex] = 2(x + 3)(x - 2)(x - 5) \\[3ex] = (2x + 6)(x^2 - 5x - 2x + 10) \\[3ex] = (2x + 6)(x^2 - 7x + 10) \\[3ex] = 2x^3 - 14x^2 + 20x + 6x^2 - 42x + 60 \\[3ex] = 2x^3 - 8x^2 - 22x + 60 \\[3ex] (b.) \\[3ex] degree = 4 \\[3ex] zeros = -2, -2, 2, 2 \\[3ex] factors = (x + 2)(x + 2)(x - 2)(x - 2) \\[3ex] factors = (x + 2)(x - 2)(x + 2)(x - 2) \\[3ex] (x + 2)(x - 2) = x^2 - 2^2 ...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \implies \\[3ex] factors = (x^2 - 2^2)(x^2 - 2^2) \\[3ex] factors = (x^2 - 4)(x^2 - 4) \\[3ex] \implies \\[3ex] f(x) = a(x^2 - 4)(x^2 - 4) \\[3ex] passes\;\;through\;\;(-3, 125) \\[3ex] To\;\;find\;\;leading\;\;coefficient:a \\[3ex] x = -3,\;\;\;f(x) = 125 \\[3ex] \implies \\[3ex] 125 = a[(-3)^2 - 4][(-3)^2 - 4] \\[3ex] 125 = a(9 - 4)(9 - 4) \\[3ex] 125 = a(5)(5) \\[3ex] 25a = 125 \\[3ex] a = \dfrac{125}{25} = 5 \\[5ex] \implies \\[3ex] f(x) = a(x^2 - 4)(x^2 - 4) \\[3ex] = 5(x^2 - 4)(x^2 - 4) \\[3ex] = (5x^2 - 20)(x^2 - 4) \\[3ex] = 5x^4 - 20x^2 - 20x^2 + 80 \\[3ex] = 5x^4 - 40x^2 + 80 $

(6.) Analyze the polynomial function: $f(x) = x^2(x - 6)$
(a.) Determine the end behavior of the graph of the function.
(b.) Find the x- and y-intercepts of the graph of the function.
(c.) Determine the zeros of the function and their multiplicity.
Does the graph touch/cross the x-axis at each zero?
(d.) Determine the maximum number of turning points on the graph of the function.
(e.) Use the above information to draw a complete graph of the function.

(a.)
$f(x) = x^2(x - 6)$
Leading term = $x^2(x) = x^3$
This implies that the graph of $f(x)$ behaves like $y = x^3$ for large values of |x|
Leading coefficient = 1 (positive)
Degree = n = 3 (odd)
End behavior is: down-up

$ (b.) \\[3ex] f(x) = x^2(x - 6) \\[3ex] \underline{x-intercept} \\[3ex] x^2(x - 6) = 0 \\[3ex] x^2 = 0 \;\;\;OR\;\;\; x - 6 = 0 \\[3ex] x = \sqrt{0} \;\;\;OR\;\;\; x = 6 \\[3ex] x = 0 \;\;\;OR\;\;\; x = 6 \\[3ex] x-intercept = (0, 0),\; (0, 0),\; (6, 0) \\[3ex] \underline{y-intercept} \\[3ex] f(0) = 0^2(0 - 6) \\[3ex] f(0) = 0(-6) \\[3ex] f(0) = 0 \\[3ex] y-intercept = (0, 0) \\[3ex] (c.) \\[3ex] \underline{1st\;\;zero} \\[3ex] x = 0 \\[3ex] Multiplicity = 2\;(even) \\[3ex] Graph\;\;touches\;\;the\;\;x-axis\;\;at\;\;x = 0 \\[3ex] \underline{2nd\;\;zero} \\[3ex] x = 6 \\[3ex] Multiplicity = 1\;(odd) \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;x = 6 \\[3ex] $ (d.) The maximum number of turning points on the graph is:
= n − 1
= 3 − 1
= 2

(e.) The graph of the function is:

Number 6

(7.) Determine the polynomial function whose
(a.) zeros are: −6, 0, 3, 1 and whose graph passes through the point $\left(-\dfrac{1}{2}, -462\right)$

(b.) zeros are: −1, 1, 3 and degree is: 3 and y-intercept is: −9

(c.) zeros are: −3, multiplicity 1; 2, multiplicity 1; 4, multiplicity 2; and degree is: 4 and contains the point (1, 72)

Type your answers in factored form.

$ (a.) \\[3ex] zeros = -6, 0, 3, 1 \\[3ex] factors = (x + 6)(x - 0)(x - 3)(x - 1) \\[3ex] \implies \\[3ex] f(x) = ax(x + 6)(x - 3)(x - 1) \\[3ex] passes\;\;through\;\;\left(-\dfrac{1}{2}, -462\right) \\[5ex] To\;\;find\;\;leading\;\;coefficient:a \\[3ex] x = -\dfrac{1}{2},\;\;\;f(x) = -462 \\[5ex] \implies \\[3ex] -462 = a * -\dfrac{1}{2} * \left(-\dfrac{1}{2} + 6\right) * \left(-\dfrac{1}{2} - 3\right) * \left(-\dfrac{1}{2} - 1\right) \\[5ex] -462 = a * -\dfrac{1}{2} * \left(-\dfrac{1}{2} + \dfrac{12}{2}\right) * \left(-\dfrac{1}{2} - \dfrac{6}{2}\right) * \left(-\dfrac{1}{2} - \dfrac{2}{2}\right) \\[5ex] -462 = a * -\dfrac{1}{2} * \dfrac{11}{2} * -\dfrac{7}{2} * -\dfrac{3}{2} \\[5ex] -462 = a * -\dfrac{231}{16} \\[5ex] a * -\dfrac{231}{16} = -462 \\[5ex] -\dfrac{16}{231} * a * -\dfrac{231}{16} = -\dfrac{16}{231} * -462 \\[5ex] a = 32 \\[3ex] \implies \\[3ex] f(x) = ax(x + 6)(x - 3)(x - 1) \\[3ex] f(x) = 32x(x + 6)(x - 3)(x - 1) \\[3ex] (b.) \\[3ex] zeros = -1, 1, 3 \\[3ex] factors = (x + 1)(x - 1)(x - 3) \\[3ex] \implies \\[3ex] f(x) = a(x + 1)(x - 1)(x - 3) \\[3ex] y-intercept = (0, -9) \\[3ex] passes\;\;through\;\;(0, -9) \\[3ex] To\;\;find\;\;leading\;\;coefficient:a \\[3ex] x = 0,\;\;\;f(x) = -9 \\[3ex] \implies \\[3ex] -9 = a(0 + 1)(0 - 1)(0 - 3) \\[3ex] -9 = a(1)(-1)(-3) \\[3ex] 3a = -9 \\[3ex] a = -\dfrac{9}{3} \\[5ex] a = -3 \\[3ex] \implies \\[3ex] f(x) = a(x + 1)(x - 1)(x - 3) \\[3ex] f(x) = -3(x + 1)(x - 1)(x - 3) \\[3ex] (c.) \\[3ex] zeros = -3, 2, 4, 4 \\[3ex] factors = (x + 3)(x - 2)(x - 4)(x - 4) \\[3ex] \implies \\[3ex] f(x) = a(x + 3)(x - 2)(x - 4)(x - 4) \\[3ex] contains\;\;the\;\;point:\;\;(1, 72) \\[3ex] To\;\;find\;\;leading\;\;coefficient:\;a \\[3ex] x = 1,\;\;\;f(x) = 72 \\[3ex] \implies \\[3ex] 72 = a(1 + 3)(1 - 2)(1 - 4)(1 - 4) \\[3ex] 72 = a(4)(-1)(-3)(-3) \\[3ex] -36a = 72 \\[3ex] a = -\dfrac{72}{36} \\[5ex] a = -2 \\[3ex] \implies \\[3ex] f(x) = a(x + 3)(x - 2)(x - 4)(x - 4) \\[3ex] f(x) -2(x + 3)(x - 2)(x - 4)^2 $

(8.) Which of the polynomial functions may have the given graph?

Number 8

$ A.\;\; f(x) = \dfrac{1}{2}(x + 2)(x + 1)(x - 1) \\[5ex] B.\;\; f(x) = \dfrac{1}{2}(x + 2)(x + 1)^2(x - 1) \\[5ex] C.\;\; f(x) = -\dfrac{1}{2}(x + 2)(x + 1)(x - 1) \\[5ex] D.\;\; f(x) = \dfrac{1}{2}(x + 2)(x + 1)(x - 1)^2 \\[5ex] E.\;\; f(x) = \dfrac{1}{2}(x + 2)^2(x + 1)(x - 1)^2 \\[5ex] F.\;\; f(x) = -\dfrac{1}{2}(x + 2)(x + 1)^2(x - 1) \\[5ex] $

Turning Points
If: Turning points: n − 1;
then Degree: at least n
Turning points = 3
n − 1 = 3
n = 3 + 1
n = 4
Degree: ≥4
Assume: degree = 4

Zeros
zeros = −2, −1, 1

Multiplicity of Zeros
The graph crosses − 2
Multiplicity of −2 = 1
The graph touches −1
Multiplicity of −1 = 2
The graph crosses 1
Multiplicity of 1 = 1

Factors
factors = (x + 2)(x + 1)²(x − 1)

Leading Coefficient
y-intercept = (0, −1)

$ f(x) = a(x + 2)(x + 1)^2(x - 1) \\[3ex] passes\;\;through\;\;y-intercept:\;(0, -1) \\[3ex] x = 0 \\[3ex] y = f(x) = f(0) = -1 \\[3ex] \implies \\[3ex] f(0) = a(0 + 2)(0 + 1)^2(0 - 1) \\[3ex] -1 = a(2)(1)^2(-1) \\[3ex] -1 = a(2)(1)(-1) \\[3ex] -2a = -1 \\[3ex] a = \dfrac{-1}{-2} \\[5ex] a = \dfrac{1}{2} \\[5ex] \implies \\[3ex] f(x) = \dfrac{1}{2}(x + 2)(x + 1)^2(x - 1) $

(9.) For these polynomial functions:

$ (I.)\;\; f(x) = -3(x - 5)(x + 3)^2 \\[3ex] (II.)\;\; f(x) = 2(x^2 + 1)^2(x - 4)^3 \\[3ex] (III.)\;\; f(x) = -8\left(x + \dfrac{1}{2}\right)^2(x + 2)^3 \\[5ex] $ (a.) List each zero and the multiplicity of each zero
(b.) Determine whether the graph crosses or touches the x-axis at each x-intercept
(c.) Determine the maximum number of turning points on the graph.
(d.) Determine the power function that the graph of f resembles for large values of |x|.
(e.) State the end behavior.

$ (I.) \\[3ex] f(x) = -3(x - 5)(x + 3)^2 \\[3ex] (a.) \\[3ex] \underline{First} \\[3ex] factor:\;\; x - 5 \\[3ex] zero:\;\; x = 5 \\[3ex] multiplicity:\;\; 1 \\[3ex] \underline{Second} \\[3ex] factor:\;\; x + 3 \\[3ex] zero:\;\; x = -3 \\[3ex] multiplicity:\;\; 2 \\[3ex] (b.) \\[3ex] \underline{First} \\[3ex] x-intercept:\;\; (5, 0) \\[3ex] multiplicity:\;\;1 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(5, 0) \\[3ex] \underline{Second} \\[3ex] x-intercept:\;\; (-3, 0) \\[3ex] multiplicity:\;\;2 = even \\[3ex] Graph\;\;touches\;\;the\;\;x-axis\;\;at\;\;(-3, 0) \\[3ex] $ Number 9I

$ (c.) \\[3ex] f(x) = -3(x - 5)(x + 3)^2 \\[3ex] f(x) = -3(x - 5)(x + 3)(x + 3) \\[3ex] Leading\;\;term = -3 \cdot x \cdot x \cdot x = -3x^3 \\[3ex] Degree:\;\;n = 3 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 3 − 1
= 2

(d.) Leading term = −3x³
Leading coefficient = −3 = negative
Degree, n = 3 = odd

(e.) Therefore, the end behavior is: up/down

$ (II.) \\[3ex] f(x) = 2(x^2 + 1)^2(x - 4)^3 \\[3ex] (a.) \\[3ex] \underline{First} \\[3ex] x^2 + 1 = 0 \\[3ex] x^2 = -1 \\[3ex] x = \sqrt{-1} = i ...imaginary\;\;number \\[3ex] No\;\;real\;\;zero \\[3ex] \underline{First} \\[3ex] factor:\;\; x - 4 \\[3ex] zero:\;\; x = 4 \\[3ex] multiplicity:\;\; 3 \\[3ex] (b.) \\[3ex] \underline{First} \\[3ex] x-intercept:\;\; (4, 0) \\[3ex] multiplicity:\;\;3 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(4, 0) \\[3ex] $ Number 9II

$ (c.) \\[3ex] f(x) = 2(x^2 + 1)^2(x - 4)^3 \\[3ex] f(x) = 2(x^2 + 1)(x^2 + 1)(x - 4)(x - 4)(x - 4) \\[3ex] Leading\;\;term = 2 \cdot x^2 \cdot x^2 \cdot x \cdot x \cdot x = 2x^7 \\[3ex] Degree:\;\;n = 7 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 7 − 1
= 6

(d.) Leading term = 2x$^7$
Leading coefficient = 2 = positive
Degree, n = 7 = odd

(e.) Therefore, the end behavior is: down/up

$ (III.) \\[3ex] f(x) = -8\left(x + \dfrac{1}{2}\right)^2(x + 2)^3 \\[5ex] (a.) \\[3ex] \underline{First} \\[3ex] factor:\;\; x + \dfrac{1}{2} \\[5ex] zero:\;\; x = -\dfrac{1}{2} \\[5ex] multiplicity:\;\; 2 \\[3ex] \underline{Second} \\[3ex] factor:\;\; x + 2 \\[3ex] zero:\;\; x = -2 \\[3ex] multiplicity:\;\; 3 \\[3ex] (b.) \\[3ex] \underline{First} \\[3ex] x-intercept:\;\; \left(-\dfrac{1}{2}, 0\right) \\[5ex] multiplicity:\;\;2 = even \\[3ex] Graph\;\;touches\;\;the\;\;x-axis\;\;at\;\;\left(-\dfrac{1}{2}, 0\right) \\[5ex] \underline{Second} \\[3ex] x-intercept:\;\; (-2, 0) \\[3ex] multiplicity:\;\;3 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(-2, 0) \\[3ex] $ Number 9III

$ (c.) \\[3ex] f(x) = -8\left(x + \dfrac{1}{2}\right)^2(x + 2)^3 \\[5ex] f(x) = -8 * \left(x + \dfrac{1}{2}\right) * \left(x + \dfrac{1}{2}\right) * (x + 2) * (x + 2) * (x + 2) \\[5ex] Leading\;\;term = -8 \cdot x \cdot x \cdot x \cdot x \cdot x = -8x^5 \\[3ex] Degree:\;\;n = 5 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 5 − 1
= 4

(d.) Leading term = −8x$^5$
Leading coefficient = −8 = negative
Degree, n = 5 = odd

(e.) Therefore, the end behavior is: up/down

(11.) Which of the polynomial functions may have the given graph?

Number 11

$ A.\;\; f(x) = x(x - 2)(x - 5) \\[3ex] B.\;\; f(x) = x(x + 2)(x + 5) \\[3ex] C.\;\; f(x) = x^2(x + 2)(x + 5) \\[3ex] D.\;\; f(x) = x^2(x - 2)(x - 5) \\[3ex] $

Turning Points
If: Turning points: n − 1;
then Degree: at least n
Turning points = 2
n − 1 = 2
n = 2 + 1
n = 3
Degree: ≥3
Assume: degree = 3

Zeros
zeros = 0, 2, 5

Multiplicity of Zeros
The graph of each zero crosses the x-axis at that zero
Hence the multiplicity of each zero = 1

Factors
factors = (x − 0)(x − 2)(x − 5)
factors = x(x − 2)(x − 5)

Leading Coefficient
y-intercept = (0, 0)...already a zero
a = 1

Therefore, $f(x) = x(x - 2)(x - 5)$

(13.) For the polynomial: f(x) = 6(x² − 9)(x² + 7x + 12):
(a.) Find the real zero(s).
(b.) Find the multiplicity of each zero.

To find the zeros of f(x); set f(x) = 0 and solve for x

$ f(x) = 6(x^2 - 9)(x^2 + 7x + 12)^2 \\[3ex] 6(x^2 - 9)(x^2 + 7x + 12)^2 = 0 \\[3ex] \implies \\[3ex] x^2 - 9 = 0 \;\;\;OR\;\;\; (x^2 + 7x + 12)^2 = 0 ...Zero\;\;Product\;\;Property \\[3ex] x^2 - 9 = 0 \\[3ex] x^2 - 3^2 = 0 \\[3ex] (x + 3)(x - 3) = 0...Difference\;\;of\;\;Two\;\;Squares \\[3ex] x + 3 = 0 \;\;\;OR\;\;\; x - 3 = 0 ...Zero\;\;Product\;\;Property \\[3ex] x = -3 \;\;\;OR\;\;\; x = 3 \\[3ex] Also: \\[3ex] (x^2 + 7x + 12)^2 = 0 \\[3ex] x^2 + 7x + 12 = \sqrt{0} \\[3ex] x^2 + 7x + 12 = 0 \\[3ex] (x + 4)(x + 3) = 0 ...Factoring\;\;Quadratic\;\;Trinomial \\[3ex] x + 4 = 0 \;\;\;OR\;\;\; x + 3 = 0 ...Zero\;\;Product\;\;Property \\[3ex] x = -4 \;\;\;OR\;\;\; x = -3...twice\;\;because\;\;of\;\;the\;\;square \\[3ex] x = -4, -3, -4, -3 \\[3ex] \implies \\[3ex] (a.) \\[3ex] real\;\;zeros: -3, 3, -4 \\[3ex] (b.) \\[3ex] multiplicity\;\;of\;\;-3 = 3 \\[3ex] multiplicity\;\;of\;\;3 = 1 \\[3ex] multiplicity\;\;of\;\;-4 = 2 $

(15.) Which of the polynomial functions may have the given graph?

Number 15

$ A.\;\; f(x) = -\dfrac{1}{8}(x + 2)^2(x - 2)(x - 4)^2 \\[5ex] B.\;\; f(x) = -\dfrac{1}{8}(x + 2)(x - 2)^2(x - 4) \\[5ex] C.\;\; f(x) = -\dfrac{1}{8}(x + 2)(x - 2)(x - 4)^2 \\[5ex] D.\;\; f(x) = -\dfrac{1}{8}(x + 2)^2(x - 2)(x - 4) \\[5ex] E.\;\; f(x) = \dfrac{1}{8}(x + 2)(x - 2)^2(x - 4) \\[5ex] F.\;\; f(x) = \dfrac{1}{8}(x + 2)(x - 2)(x - 4) \\[5ex] $

Turning Points
If: Turning points: n − 1;
then Degree: at least n
Turning points = 3
n − 1 = 3
n = 3 + 1
n = 4
Degree: ≥4
Assume: degree = 4

Zeros
zeros = −2, 2, 4

Multiplicity of Zeros
The graph crosses − 2
Multiplicity of −2 = 1
The graph touches 2
Multiplicity of 2 = 2
The graph crosses 4
Multiplicity of 4 = 1

Factors
factors = (x + 2)(x − 2)²(x − 4)

Leading Coefficient
y-intercept = (0, 4)

$ f(x) = a(x + 2)(x - 2)^2(x - 4) \\[3ex] passes\;\;through\;\;y-intercept:\;(0, 4) \\[3ex] x = 0 \\[3ex] y = f(x) = f(0) = 4 \\[3ex] \implies \\[3ex] f(0) = a(0 + 2)(0 - 2)^2(0 - 4) \\[3ex] 4 = a(2)(-2)^2(-4) \\[3ex] 4 = a(2)(4)(-4) \\[3ex] -32a = 4 \\[3ex] a = \dfrac{4}{-32} \\[5ex] a = -\dfrac{1}{8} \\[5ex] \implies \\[3ex] f(x) = -\dfrac{1}{8}(x + 2)(x - 2)^2(x - 4) $

(19.) For each polynomial graph:
(a.) Determine whether the graph could be graph of a polynomial function.
If it is the graph of a polynomial function:
(b.) List the real zeros.
(c.) State the least degree the polynomial function can have.

(A.) Number 19A

(B.)

(A.)
(a.) The graph is smooth and continuous. It is the graph of a polynomial function.
(b.) The real zeros are: −2, 1, and 3
(c.) If: Turning points: n − 1;
then Degree: at least n
Turning points = 2
n − 1 = 2
n = 2 + 1
n = 3
Degree: ≥3

(B.)
The graph has a gap. It is not continuous. It is not the graph of a polynomial function.

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(23.) The coordinates of the indicated point in the graph is $\left(-1, -\dfrac{45}{16}\right)$
Using the smallest possible degree, which of the polynomial functions may have the given graph?

Number 23

$ A.\;\; f(x) = \dfrac{1}{16}(x + 2)(x - 2)(x - 4) \\[5ex] B.\;\; f(x) = \dfrac{1}{16}(x + 2)^2(x - 2)(x - 4) \\[5ex] C.\;\; f(x) = -\dfrac{1}{16}(x + 2)(x - 2)^2(x - 4) \\[5ex] D.\;\; f(x) = \dfrac{1}{16}(x + 2)(x - 2)(x - 4)^2 \\[5ex] E.\;\; f(x) = \dfrac{1}{16}(x + 2)(x - 2)^2(x - 4) \\[5ex] F.\;\; f(x) = -\dfrac{1}{16}(x + 2)(x - 2)(x - 4)^2 \\[5ex] $

Turning Points
If: Turning points: n − 1;
then Degree: at least n
Turning points = 3
n − 1 = 3
n = 3 + 1
n = 4
Degree: ≥4
Smallest degree = 4

Zeros
zeros = −2, 2, 4

Multiplicity of Zeros
The graph crosses − 2
Multiplicity of −2 = 1
The graph touches 2
Multiplicity of 2 = 2
The graph crosses 4
Multiplicity of 4 = 1

Factors
factors = (x + 2)(x − 2)²(x − 4)

Leading Coefficient
Point = $\left(-1, -\dfrac{45}{16}\right)$

$ f(x) = a(x + 2)(x - 2)^2(x - 4) \\[3ex] passes\;\;through\;\;\left(-1, -\dfrac{45}{16}\right) \\[5ex] x = -1 \\[3ex] y = f(x) = f(-1) = -\dfrac{45}{16} \\[5ex] \implies \\[3ex] f(-1) = a(-1 + 2)(-1 - 2)^2(-1 - 4) \\[5ex] -\dfrac{45}{16} = a(1)(-3)^2(-5) \\[5ex] -\dfrac{45}{16} = a(9)(-5) \\[5ex] -45a = -\dfrac{45}{16} \\[5ex] -\dfrac{1}{45} * -45 * a = -\dfrac{1}{45} * -\dfrac{45}{16} \\[5ex] a = \dfrac{1}{16} \\[5ex] \implies \\[3ex] f(x) = \dfrac{1}{16}(x + 2)(x - 2)^2(x - 4) \\[5ex] (x + 2)(x - 2)^2(x - 4) \\[3ex] = (x + 2)(x - 2)(x - 2)(x - 4) \\[3ex] = (x^2 - 2^2)(x^2 - 4x - 2x + 8) \\[3ex] = (x^2 - 4)(x^2 - 6x + 8) \\[3ex] = x^4 - 6x^3 + 8x^2 - 4x^2 + 24x - 32 \\[3ex] = x^4 - 6x^3 + 4x^2 + 24x - 32 \\[3ex] \implies \\[3ex] f(x) = \dfrac{1}{16}\left(x^4 - 6x^3 + 4x^2 + 24x - 32\right) \\[5ex] = \dfrac{x^4}{16} - \dfrac{6x^3}{16} + \dfrac{4x^2}{16} + \dfrac{24x}{16} - \dfrac{32}{16} \\[5ex] = \dfrac{x^4}{16} - \dfrac{3x^3}{8} + \dfrac{x^2}{4} + \dfrac{3x}{2} - 2 $

(29.) The coordinates of the indicated point in the graph is (−1, −16)
Using the smallest possible degree, which of the polynomial functions may have the given graph?

Number 29

$ A.\;\; f(x) = x(x - 2)^2(x - 3)^2 \\[3ex] B.\;\; f(x) = x(x + 2)^2(x + 3)^2 \\[3ex] C.\;\; f(x) = -x(x + 2)(x - 3) \\[3ex] D.\;\; f(x) = -x(x + 2)^2(x - 3)^2 \\[3ex] E.\;\; f(x) = x(x + 2)(x - 3) \\[3ex] F.\;\; f(x) = x(x + 2)^2(x - 3)^2 \\[3ex] $

Turning Points
If: Turning points: n − 1;
then Degree: at least n
Turning points = 4
n − 1 = 4
n = 4 + 1
n = 5
Degree: ≥5
Smallest degree = 5

Zeros
zeros = −2, 0, 3

Multiplicity of Zeros
The graph touches − 2
Multiplicity of −2 = 2
The graph crosses 0
Multiplicity of 0 = 1
The graph touches 3
Multiplicity of 3 = 2

Factors
factors = (x + 2)^2(x − 0)(x − 3)^2
factors = x(x + 2)^2(x − 3)^2

Leading Coefficient
Point = (−1, −16)

$ f(x) = ax(x + 2)^2(x - 3)^2\\[3ex] passes\;\;through\;\;(-1, -16) \\[5ex] x = -1 \\[3ex] y = f(x) = f(-1) = -16 \\[3ex] \implies \\[3ex] f(-1) = a(-1)(-1 + 2)^2(-1 - 3)^2 \\[3ex] -16 = a(-1)(1)^2(-4)^2 \\[3ex] -16 = -a(1)(16) \\[3ex] -16 = -16a \\[3ex] -16a = -16 \\[3ex] a = \dfrac{-16}{-16} \\[5ex] a = 1 \\[3ex] \implies \\[3ex] f(x) = 1x(x + 2)^2(x - 3)^2 \\[3ex] f(x) = x(x + 2)^2(x - 3)^2 $

(39.) For these polynomial functions:

$ (I.)\;\; f(x) = (x - 7)^3(x + 6)^2 \\[3ex] (II.)\;\; f(x) = \dfrac{1}{3}(3x^2 + 8)^2(x^2 + 9) \\[5ex] (III.)\;\; f(x) = -4x^2(x^2 - 7) \\[3ex] $ (a.) List each zero and the multiplicity of each zero
(b.) Determine whether the graph crosses or touches the x-axis at each x-intercept
(c.) Determine the maximum number of turning points on the graph.
(d.) Determine the power function that the graph of f resembles for large values of |x|.
(e.) State the end behavior.

$ (I.) \\[3ex] f(x) = (x - 7)^3(x + 6)^2 \\[3ex] (a.) \\[3ex] \underline{First} \\[3ex] factor:\;\; x - 7 \\[3ex] zero:\;\; x = 7 \\[3ex] multiplicity:\;\; 3 \\[3ex] \underline{Second} \\[3ex] factor:\;\; x + 6 \\[3ex] zero:\;\; x = -6 \\[3ex] multiplicity:\;\; 2 \\[3ex] (b.) \\[3ex] \underline{First} \\[3ex] x-intercept:\;\; (7, 0) \\[3ex] multiplicity:\;\;3 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(7, 0) \\[3ex] \underline{Second} \\[3ex] x-intercept:\;\; (-6, 0) \\[3ex] multiplicity:\;\;2 = even \\[3ex] Graph\;\;touches\;\;the\;\;x-axis\;\;at\;\;(-6, 0) \\[3ex] $ Number 39I

$ (c.) \\[3ex] f(x) = (x - 7)^3(x + 6)^2 \\[3ex] f(x) = (x - 7)(x - 7)(x - 7)(x + 6)(x + 6) \\[3ex] Leading\;\;term = x \cdot x \cdot x \cdot x \cdot x = x^5 \\[3ex] Degree:\;\;n = 5 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 5 − 1
= 4

(d.) Leading term = x$^5$
Leading coefficient = 1 = positive
Degree, n = 5 = odd

(e.) Therefore, the end behavior is: down/up

$ (II.) \\[3ex] f(x) = \dfrac{1}{3}(3x^2 + 8)^2(x^2 + 9) \\[5ex] (a.) \\[3ex] \underline{First} \\[3ex] 3x^2 + 8 = 0 \\[3ex] 3x^2 = -8 \\[3ex] x^2 = -\dfrac{8}{3} \\[5ex] x = \sqrt{\left(-\dfrac{8}{3}\right)} ...imaginary\;\;number \\[5ex] No\;\;real\;\;zero \\[3ex] \underline{First} \\[3ex] x^2 + 9 = 0 \\[3ex] x^2 = -9 \\[3ex] x = \sqrt{-9} = 3i ...imaginary\;\;number \\[3ex] No\;\;real\;\;zero \\[3ex] $ (b.) The graph neither crosses nor touches the x-axis

Number 39II

$ (c.) \\[3ex] f(x) = \dfrac{1}{3}(3x^2 + 8)^2(x^2 + 9) \\[5ex] f(x) = \dfrac{1}{3}(3x^2 + 8)(3x^2 + 8)(x^2 + 9) \\[5ex] Leading\;\;term = \dfrac{1}{3} \cdot 3x^2 \cdot 3x^2 \cdot x^2 = 3x^6 \\[3ex] Degree:\;\;n = 6 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 6 − 1
= 5

(d.) Leading term = 3x$^6$
Leading coefficient = 3 = positive
Degree, n = 6 = even

(e.) Therefore, the end behavior is: up/up

$ (III.) \\[3ex] f(x) = -4x^2(x^2 - 7) \\[3ex] (a.) \\[3ex] \underline{First} \\[3ex] factor:\;\; x + 0 \\[5ex] zero:\;\; x = 0 \\[5ex] multiplicity:\;\; 2 \\[3ex] \underline{Second} \\[3ex] x^2 - 7 = 0 \\[3ex] x^2 = 7 \\[3ex] x = \pm\sqrt{7} \\[3ex] factor:\;\; x - \sqrt{7} \\[3ex] zero:\;\; x = \sqrt{7} \\[3ex] multiplicity:\;\; 1 \\[3ex] \underline{Third} \\[3ex] factor:\;\; x + \sqrt{7} \\[3ex] zero:\;\; x = -\sqrt{7} \\[3ex] multiplicity:\;\; 1 \\[3ex] (b.) \\[3ex] \underline{First} \\[3ex] x-intercept:\;\; (0, 0) \\[3ex] multiplicity:\;\;2 = even \\[3ex] Graph\;\;touches\;\;the\;\;x-axis\;\;at\;\;(0, 0) \\[3ex] \underline{Second} \\[3ex] x-intercept:\;\; (\sqrt{7}, 0) \\[3ex] multiplicity:\;\;1 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(\sqrt{7}, 0) \\[3ex] \underline{Third} \\[3ex] x-intercept:\;\; (-\sqrt{7}, 0) \\[3ex] multiplicity:\;\;1 = odd \\[3ex] Graph\;\;crosses\;\;the\;\;x-axis\;\;at\;\;(-\sqrt{7}, 0) \\[3ex] $ Number 39III

$ (c.) \\[3ex] f(x) = -4x^2(x^2 - 7) \\[3ex] f(x) = -4 \cdot x^2 \cdot (x^2 - 7) \\[3ex] Leading\;\;term = -4 \cdot x^2 \cdot x^2 = -4x^4 \\[3ex] Degree:\;\;n = 4 \\[3ex] $ The maximum number of turning points on the graph is:
= n − 1
= 4 − 1
= 3

(d.) Leading term = −4x$^4$
Leading coefficient = −4 = negative
Degree, n = 4 = even

(e.) Therefore, the end behavior is: down/down

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