Polynomial Functions and Their Graphs: Algebra 2 Math Worksheet Answer Key – PDF Download

Algebra 2 Polynomial Worksheet Answer Key: problems 1–5

Problem 1: Identify polynomials

1. Yes: whole-number exponents \(4\) and \(1\).
2. No: \(\frac{2}{x}=2x^{-1}\) has a negative exponent.
3. No: \(\sqrt{x}=x^{1/2}\) has a fractional exponent.
4. Yes: a constant is a polynomial of degree \(0\).

Problem 2: Degree and leading coefficient

1. Degree \(3\), leading coefficient \(4\).
2. Degree \(5\), leading coefficient \(-2\).
3. Three linear factors give degree \(3\); the leading coefficient is \(1\cdot 1\cdot 2=\) \(2\).
4. \(p(x)=3x^{2}-x^{3}\): degree \(3\), leading coefficient \(-1\).

Problem 3: Evaluate a polynomial

\(f(-2)=-8-8-2-5=\) \(-23\).

\(f(0)=\) \(-5\).

\(f(3)=27-18+3-5=\) \(7\).

Problem 4: End behavior

1. Even degree, positive leading coefficient: rises on both ends.
2. Odd degree, negative leading coefficient: rises on the left, falls on the right.
3. Even degree, negative leading coefficient: falls on both ends.
4. Odd degree, positive leading coefficient: falls on the left, rises on the right.

Problem 5: Zeros from factored form

1. \(x=4\) and \(x=-1\).
2. \(x=0\), \(5\) and \(-3\).
3. \(2x-6=0\) gives \(x=3\); the other zero is \(x=-7\).
4. \(x=0\), \(2\) and \(-6\).

Algebra 2 Polynomial Worksheet Answer Key: problems 6–10

Problem 6: Multiplicity

1. \(x=0\) (multiplicity \(1\), crosses), \(x=-3\) (multiplicity \(2\), touches), \(x=1\) (multiplicity \(3\), flattens and crosses). Degree \(1+2+3=\) \(6\).
2. \(x=5\) (multiplicity \(4\), touches) and \(x=-2\) (multiplicity \(1\), crosses). Degree \(4+1=\) \(5\).

Problem 7: Factor to find zeros

1. \(x(x-3)(x+3)\): zeros \(0,\ 3,\ -3\).
2. \(x(x^{2}+2x-8)=x(x+4)(x-2)\): zeros \(0,\ -4,\ 2\).
3. \(x^{2}(x-4)(x+4)\): zeros \(0\) (multiplicity 2), \(4,\ -4\).
4. Group: \(x^{2}(x-5)-4(x-5)=(x-5)(x^{2}-4)=(x-5)(x-2)(x+2)\): zeros \(5,\ 2,\ -2\).

Problem 8: Write a polynomial from its zeros

1. \((x+2)(x-1)=x^{2}+x-2\), and \((x^{2}+x-2)(x-4)=x^{3}-4x^{2}+x^{2}-4x-2x+8\). \(f(x)=x^{3}-3x^{2}-6x+8\)
2. \(2x(x-3)^{2}=2x(x^{2}-6x+9)\). \(f(x)=2x^{3}-12x^{2}+18x\)

Problem 9: Find the y-intercept

1. \(f(0)=2(-3)(1)(-4)=\) \(24\).
2. \(g(0)=-(4)(-5)=\) \(20\).
3. \(h(0)=\) \(6\), the constant term.

Problem 10: Where is the function positive?

The zeros \(-3\), \(1\) and \(4\) split the line into four intervals. Test points: \(f(-4)=-40\), \(f(0)=12\), \(f(2)=-10\), \(f(5)=32\).

\(f(x)>0\) on \((-3, 1)\) and \((4, \infty)\); \(f(x)< 0\) on \((-\infty, -3)\) and \((1, 4)\).

Algebra 2 Polynomial Worksheet Answer Key: problems 11–15

Problem 11: Turning points

1. 3 turning points (two relative minimums and one relative maximum).
2. At most \(n-1\) turning points, so \(n-1\ge 3\) and the least degree is 4.
3. Both ends rise, so the degree is even and the leading coefficient is positive.
4. \(7-1=\) \(6\).

Problem 12: Equation from a graph

The zero at \(1\) is a touch point, so its multiplicity is even: \(2\). Write \(f(x)=a(x+2)(x-1)^{2}(x-3)\). Use \(f(0)=-3\): \(a(2)(1)(-3)=-6a=-3\), so \(a=\frac{1}{2}\).

\(f(x)=\frac{1}{2}(x+2)(x-1)^{2}(x-3)\). Check: \(f(2)=\frac{1}{2}(4)(1)(-1)=-2\), a point below the axis, as the graph shows.

Problem 13: Sketch a cubic

Zeros: \(-3\), \(0\) and \(2\), all multiplicity \(1\), so the graph crosses at each. The y-intercept is \(f(0)=0\). Degree \(3\) with positive leading coefficient: falls on the left, rises on the right. Test points: \(f(-2)=8\) and \(f(1)=-4\), so the graph is above the axis between \(-3\) and \(0\) and below it between \(0\) and \(2\).

Problem 14: Sketch with a double zero

Zeros: \(-1\) (multiplicity \(2\), touches) and \(2\) (multiplicity \(1\), crosses). Degree \(3\) with negative leading coefficient: rises on the left, falls on the right. The y-intercept is \(g(0)=-(1)(-2)=2\). Also \(g(1)=-(4)(-1)=4\).

Problem 15: Degree 5 with a negative leading coefficient

Odd degree with negative leading coefficient: rises on the left and falls on the right. At most \(5-1=\) 4 turning points and at most 5 real zeros. Because the ends go in opposite directions, the graph must cross the x-axis at least once: at least 1 real zero.

Algebra 2 Polynomial Worksheet Answer Key: problems 16–20

Problem 16: Factor a quartic

Treat \(x^{2}\) as one unit: \(x^{4}-5x^{2}+4=(x^{2}-1)(x^{2}-4)=(x-1)(x+1)(x-2)(x+2)\).

Zeros: \(-2,\ -1,\ 1,\ 2\). Y-intercept: \(f(0)=4\). Even degree, positive leading coefficient: rises on both ends.

Problem 17: Equation from zeros and a point

\(f(x)=a(x-2)^{2}(x+3)\). Then \(f(0)=a(4)(3)=12a=-24\), so \(a=-2\).

\(f(x)=-2(x-2)^{2}(x+3)\). Check: \(f(1)=-2(1)(4)=-8\) and \(f(0)=-2(4)(3)=-24\).

Problem 18: Read a table

1. \(f(x)=0\) at \(x=-2\) and \(x=1\).
2. Increasing from \(x=-3\) to \(x=-1\) (values \(-16, 0, 4\)); decreasing from \(x=-1\) to \(x=1\) (\(4, 2, 0\)); increasing from \(x=1\) to \(x=3\) (\(0, 4, 20\)). Increasing on \((-3,-1)\) and \((1,3)\); decreasing on \((-1,1)\).
3. The values rise to \(4\) and then fall: relative maximum \((-1, 4)\). They fall to \(0\) and then rise: relative minimum \((1, 0)\).

Problem 19: Open-top box

1. The width \(10-2x\) must be positive, so \(0 < x < 5\).
2. \(V(1)=1(8)(14)=112\), \(V(2)=2(6)(12)=144\), \(V(3)=3(4)(10)=120\), \(V(4)=4(2)(8)=64\). \(112,\ 144,\ 120,\ 64\)
3. \(x=2\) inches, with a volume of \(144\) cubic inches.
4. \(x=0,\ 5,\ 8\). Only \(0\) and \(5\) bound the domain; at \(x=8\) the dimensions would be negative.

Problem 20: Shipping crate

1. \(x(x^{2}+2x-8)=\) \(x^{3}+2x^{2}-8x\).
2. The height \(x-2\) must be positive, so \(x>2\); at \(x=2\) the volume is \(0\).
3. \(V(4)=4(8)(2)=\) \(64\) cubic feet.

Algebra 2 Polynomial Worksheet Answer Key: problems 21–24

Problem 21: Hiking trail

1. \(E(x)=0\) at \(0,\ 3\) and \(6\) miles.
2. \(E(4)=4(1)(-2)=-8 < 0\), so between \(3\) and \(6\) miles.
3. \(7(4)(1)=\) \(28\) feet above the trailhead.
4. \(E(1)=10\), \(E(2)=8\), \(E(4)=-8\), \(E(5)=-10\). The lowest whole-mile point is mile \(5\) at \(-10\) feet.

Problem 22: Bake sale profit

1. \(x=0\), \(10\) and \(30\) dozen: \(10\) and \(30\) dozen are the real break-even amounts (the zero at \(0\) means no baking).
2. \(P(15)=\$562.50\), \(P(20)=\$1{,}000\), \(P(22)=-0.5(22)(12)(-8)=\$1{,}056\), \(P(25)=-0.5(25)(15)(-5)=\$937.50\).
3. \(22\) dozen, with a profit of \$1,056.
4. \(P(5)=-0.5(5)(-5)(-25)=-312.5\), so she loses \$312.50.

Problem 23: Drone flight

1. \(d(4)=4(4)=\) \(16\) feet; \(d(5)=5(1)=\) \(5\) feet.
2. \(d(1)=25\), \(d(2)=32\), \(d(3)=27\), \(d(4)=16\), \(d(5)=5\). The greatest is \(32\) feet at \(2\) seconds.
3. The even multiplicity means the graph touches zero without crossing: the drone comes down gently and just reaches the ground at \(6\) seconds.

Problem 24: Skate ramp

1. \(h(4)=a(4)(-8)^{2}=256a=8\), so \(a=\frac{1}{32}\). \(h(x)=\frac{1}{32}x(x-12)^{2}\)
2. \(h(6)=\frac{6(36)}{32}=\) \(6.75\) feet.
3. \(h(3)=\frac{243}{32}\approx 7.59\), \(h(4)=8\), \(h(5)=\frac{245}{32}\approx 7.66\). The ramp is highest at \(x=4\) feet, \(8\) feet tall.

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