Free math worksheets

- Compare \(f(-x)\) with \(f(x)\). - An even power is unchanged when \(x\) is replaced by \(-x\).

1. Substitute \(-x\): \(f(-x)=(-x)^4+3=x^4+3\). 2. Since \(f(-x)=f(x)\), the function is even.

Even.

- Compare the outputs at opposite inputs. - Check whether the domain contains both \(x\) and \(-x\) for each nonzero input.

1. The domain is symmetric about \(0\). 2. The paired inputs have matching outputs: \(h(-1)=h(1)=2\) and \(h(-3)=h(3)=4\). 3. Therefore, \(h(-x)=h(x)\) for every input in the domain, so \(h\) is even.

Even.

- Recall the equation that defines an odd function. - Expand each expression and examine the powers of \(x\). - You can also compare the function values at \(x = 1\) and \(x = -1\).

1. A graph is symmetric about the origin when its function is odd, so \(f(-x) = -f(x)\). 2. For a), \(f(x) = x(4 - x^2) = 4x - x^3\). Both terms have odd powers, so \(f(-x) = -f(x)\). 3. For b), \(f(x) = 4x - x^2\), which contains both odd and even powers. 4. For c), \(f(x) = 16 - x^2\), which is even and is symmetric about the y-axis. 5. For d), \(f(x) = 16x - 8x^2 + x^3\), which contains both odd and even powers.

a) \(y = x(4 - x^2)\)

- Use the condition \(f(-x)=f(x)\). - Consider how even and odd powers change when \(x\) is replaced by \(-x\). - Expanding or using a special-product identity may reveal the powers more clearly.

1. A graph is symmetric about the y-axis when \(f(-x)=f(x)\). 2. For choice d), use the difference of squares: \(f(x)=x^6-1\). 3. Then \(f(-x)=(-x)^6-1=x^6-1=f(x)\), so choice d) is even. 4. The other choices contain a nonzero odd-power term after expansion, so they do not satisfy \(f(-x)=f(x)\).

d) \(f(x)=(x^3-1)(x^3+1)\)

- Apply \(f(-x)\) to each candidate before using either given point. - Compare how y-axis symmetry and origin symmetry transform the point \((2,-16)\). - Do not recompute the formulas at \(x=-2\); use the established parity instead.

1. For \(p\), \(p(-x)=-(-x)^4=-x^4=p(x)\), so \(p\) is even. 2. For \(q\), \(q(-x)=-2(-x)^3=2x^3=-q(x)\), so \(q\) is odd. 3. An even function sends opposite inputs to the same output, so because \(p(2)=-16\), symmetry would require \(p(-2)=-16\), not \(16\). 4. An odd function sends opposite inputs to opposite outputs, so because \(q(2)=-16\), symmetry requires \(q(-2)=16\). Therefore, \(q\) is the candidate that passes through \(R\).

a) \(p\) is even and \(q\) is odd. b) \(q(x)=-2x^3\), because odd symmetry sends \((2,-16)\) to \((-2,16)\).

- Classify each power function as even or odd before considering \(-r\). - For an even function, opposite inputs have the same output. - For an odd function, opposite inputs have opposite outputs.

1. The function \(x^6\) is even, so \((-r)^6=r^6\). Therefore, if \(r^6=10\), then \((-r)^6=10\); \(-r\) solves the same equation. 2. The function \(x^3\) is odd, so \((-r)^3=-r^3\). Since \(r^3=-8\), \((-r)^3=8\); \(-r\) solves the sign-changed equation. 3. The function \(x^4\) is even, so \((-r)^4=r^4=5\); \(-r\) solves the same equation. 4. The function \(x^5\) is odd, so \((-r)^5=-r^5=-7\); \(-r\) solves the sign-changed equation.

a) Same equation: \((-r)^6=10\) b) Sign-changed equation: \((-r)^3=8\) c) Same equation: \((-r)^4=5\) d) Sign-changed equation: \((-r)^5=-7\)

- The only algebraic difference between the two cases is the sign of \((-r)^n\). - Connect the resulting equality to the definitions of even and odd functions. - Translate the input-output relationship into a statement about reflected points on the graph.

1. If \(n\) is even, then \((-r)^n=r^n\), so \(p_n(-r)=p_n(r)\). Thus, whenever \(p_n(r)=c\ne0\), the opposite input \(-r\) produces the same output \(c\). 2. If \(n\) is odd, then \((-r)^n=-r^n\), so \(p_n(-r)=-p_n(r)\). Thus, if \((r, c)\) lies on the graph, then \((-r, -c)\) also lies on the graph. 3. Even-power graphs are symmetric about the y-axis. Odd-power graphs are symmetric about the origin.

a) For even \(n\), \(p_n(-r)=p_n(r)\); opposite inputs give the same output. b) For odd \(n\), \(p_n(-r)=-p_n(r)\); \((r, c)\) pairs with \((-r, -c)\). c) Even \(n\): y-axis symmetry. Odd \(n\): origin symmetry.

- Rewrite the claim as a statement comparing \(f(r)\) and \(f(-r)\). - Use the defining equation for an even function in one case and an odd function in the other. - Do not calculate any specific roots; the statement is about all inputs.

1. If \(n\) is even, then \(f(x)=x^n\) is even, so \(f(-r)=f(r)=c\). Therefore, \(-r\) solves \(x^n=c\), not \(x^n=-c\) when \(c\ne0\). The claim is false for even exponents. 2. If \(n\) is odd, then \(f(x)=x^n\) is odd, so \(f(-r)=-f(r)=-c\). Therefore, \(-r\) does solve \(x^n=-c\). The claim is true for odd exponents.

Even \(n\): false, because \((-r)^n=r^n=c\). Odd \(n\): true, because \((-r)^n=-r^n=-c\).

- Determine how \((-x)^n\) behaves for even and odd \(n\). - Apply that sign behavior inside the full expression \(kx^n\). - Use the appropriate graph symmetry to infer the point at the opposite input.

1. If \(n\) is even, then \((-x)^n=x^n\). Therefore, \(f(-x)=k(-x)^n=kx^n=f(x)\), so \(f\) is even. Since \((-2, 4)\) is on the graph, \((2, 4)\) is also on the graph. 2. If \(n\) is odd, then \((-x)^n=-x^n\). Therefore, \(f(-x)=k(-x)^n=-kx^n=-f(x)\), so \(f\) is odd. Since \((-2, 4)\) is on the graph, \((2, -4)\) is also on the graph. 3. The constant multiplier \(k\) appears in both \(f(x)\) and \(f(-x)\). It does not change whether replacing \(x\) by \(-x\) preserves the output or changes its sign, so it does not change the parity determined by \(n\).

a) \((2, 4)\); for even \(n\), \(f\) is even. b) \((2, -4)\); for odd \(n\), \(f\) is odd. c) Multiplication by nonzero \(k\) preserves the relation \(f(-x)=f(x)\) for even \(n\) and \(f(-x)=-f(x)\) for odd \(n\).

- Replace \(x\) with \(-x\) in each rule. - Compare the result with the original function and its negative. - Expand the product in part c) if needed.

1. For \(f\), \(f(-x)=\frac{(-x)^2-9}{(-x)^2+1}=\frac{x^2-9}{x^2+1}=f(x)\). Therefore, \(f\) is even. 2. For \(g\), \(g(-x)=\frac{1}{2}(-x)^3-2(-x)=-\frac{1}{2}x^3+2x=-g(x)\). Therefore, \(g\) is odd. 3. Expand \(h(x)=x^3-3x^2\). Then \(h(-x)=-x^3-3x^2\), which is neither \(h(x)\) nor \(-h(x)\). Therefore, \(h\) is neither even nor odd.

a) Even; symmetric about the y-axis b) Odd; symmetric about the origin c) Neither even nor odd

- Identify the term that prevents the polynomial from being even. - Set its coefficient equal to \(0\). - Solve the resulting quadratic equation completely.

1. An even polynomial can contain only even powers with nonzero coefficients. 2. Therefore, the coefficient of \(x^3\) must equal \(0\): \(a^2-1=0\). 3. Factoring gives \((a-1)(a+1)=0\), so \(a=1\) or \(a=-1\). 4. Thus, more than one real parameter value works.

There are two values: \(a=-1\) and \(a=1\).

- Classify each exponent as even or odd. - Remember that a nonzero constant is an \(x^0\) term. - Decide whether changing one coefficient affects the other terms.

1. Every nonzero term has an even exponent, including the constant term with exponent \(0\). Therefore, \(f(-x)=f(x)\), so \(f\) is even. 2. An odd polynomial can contain only odd-power terms with nonzero coefficients. Changing the constant term can remove the \(x^0\) term by setting it to \(0\), but the terms \(3x^4\) and \(-5x^2\) remain. 3. Therefore, changing only the constant term cannot make the function odd.

1) The function is even and its graph is symmetric about the y-axis. 2) It is impossible. The even-power terms \(3x^4\) and \(-5x^2\) remain regardless of the constant term.

- Replace \(x\) with \(-x\) in every term. - Compare the shifted function at \(-x\) with its negative. - Interpret adding \(3\) as a vertical translation.

1. Substitute \(-x\): \(h(-x)=(-x)^5-4(-x)^3+k(-x)=-x^5+4x^3-kx=-h(x)\). Thus, \(h\) is odd for every real \(k\). 2. For \(g\), \(g(-x)=-h(x)+3\), while \(-g(x)=-h(x)-3\). Since these expressions are not equal, \(g\) is not odd. 3. Geometrically, adding \(3\) shifts the graph upward, moving its center of symmetry from \((0, 0)\) to \((0, 3)\).

1) \(h(-x)=-h(x)\) for every \(k\), so \(h\) is odd. 2) \(g(-x)=-h(x)+3\ne-h(x)-3=-g(x)\), so \(g\) is not odd.

- Find the values that make the denominator \(0\). - Check whether every excluded value has its opposite excluded as well. - Both standard symmetries require a domain that is symmetric about \(0\).

1. The denominator is \(0\) at \(x=2\) and \(x=-5\), so the domain is \(\mathbb{R}\setminus\{-5, 2\}\). 2. Either y-axis symmetry or origin symmetry requires the domain to be symmetric about \(0\): whenever \(x\) is in the domain, \(-x\) must also be in the domain. 3. Here, \(5\) is in the domain but \(-5\) is not. Similarly, \(-2\) is in the domain but \(2\) is not. 4. Therefore, the domain fails a necessary condition for both symmetries.

The domain is \(\mathbb{R}\setminus\{-5, 2\}\), which is not symmetric about \(0\). Therefore, the graph has neither y-axis symmetry nor origin symmetry.

- Numerical values at opposite inputs can suggest a pattern, but they do not prove a function is even or odd. - Substitute \(-x\) for the input and simplify before comparing with the original function. - Compare the simplified result with \(f(x)\) and with \(-f(x)\).

1. a) \(T_1(4)=3\cdot4^2+10=58\) and \(T_1(-4)=3\cdot(-4)^2+10=58\). Also, \(T_2(4)=2\cdot4^3=128\) and \(T_2(-4)=2\cdot(-4)^3=-128\). 2. b) \(T_1(-x)=3(-x)^2+10=3x^2+10=T_1(x)\). 3. \(T_2(-x)=2(-x)^3=-2x^3=-T_2(x)\). 4. c) Because \(T_1(-x)=T_1(x)\) for every real \(x\), \(T_1\) is even. Because \(T_2(-x)=-T_2(x)\) for every real \(x\), \(T_2\) is odd.

a) \(T_1(4)=T_1(-4)=58\); \(T_2(4)=128\) and \(T_2(-4)=-128\) b) \(T_1(-x)=T_1(x)\); \(T_2(-x)=-T_2(x)\) c) \(T_1\) is even and \(T_2\) is odd, by the defining relationships in part b.

- Determine whether each symmetry condition requires even powers or odd powers. - Treat the constant \(c\) as the term \(cx^0\). - Decide whether a vertical shift is compatible with symmetry about the origin.

1. For symmetry about the y-axis, \(f\) must be even. Therefore, \(n\) and \(m\) must both be even. The constant \(c\) may be any real number. Two examples are \((n, m, c) = (2, 4, 3)\) and \((6, 2, 0)\). 2. For symmetry about the origin, \(f\) must be odd. Therefore, \(n\) and \(m\) must both be odd, and the constant term must be zero. Two examples are \((n, m, c) = (1, 3, 0)\) and \((5, 1, 0)\).

a) Sample answers: \((n, m, c) = (2, 4, 3)\) and \((6, 2, 0)\). In general, \(n, m \in \{2, 4, 6\}\) and \(c \in \mathbb{R}\). b) Sample answers: \((n, m, c) = (1, 3, 0)\) and \((5, 1, 0)\). In general, \(n, m \in \{1, 3, 5\}\) and \(c = 0\).

- Substitute \(-x\) for every occurrence of \(x\). - Simplify powers with even exponents. - Compare the result with \(f(x)\) and \(-f(x)\).

1. Substitute \(-x\) for \(x\): \(f(-x) = 3(-x)^6 - 4(-x)^2 + 7\). 2. Because the exponents \(6\) and \(2\) are even, \(f(-x) = 3x^6 - 4x^2 + 7\). 3. Therefore, \(f(-x) = f(x)\), so \(f\) is even and its graph is symmetric about the y-axis.

The graph is symmetric about the y-axis because \(f(-x) = f(x)\).

- Substitute \(-x\) for every occurrence of \(x\). - Simplify one power at a time. - Compare the result with the original function. - If the result is not the original function, compare it with the opposite of the function.

1. \(p(-x) = 3(-x)^5 - 2(-x)^3 + (-x) = -3x^5 + 2x^3 - x = -p(x)\). Therefore, \(p\) is odd. 2. \(q(-x) = (-x)^4 - 2(-x)^2 + 5 = x^4 - 2x^2 + 5 = q(x)\). Therefore, \(q\) is even.

1. \(p\) is odd; its graph is symmetric about the origin. 2. \(q\) is even; its graph is symmetric about the y-axis.

- Replace \(x\) with \(-x\) in each function. - Recall the symmetry identity for cosine. - Compare the result with both \(f(x)\) and \(-f(x)\).

1. For \(f\), substitute \(-x\): \(f(-x)=\frac{e^{-x}+e^x}{2}=f(x)\). Therefore, \(f\) is even, and its graph is symmetric about the y-axis. 2. For \(g\), \(g(-x)=(-x)^3\cos(-x)\). Since \((-x)^3=-x^3\) and \(\cos(-x)=\cos(x)\), \(g(-x)=-x^3\cos(x)=-g(x)\). Therefore, \(g\) is odd, and its graph has origin symmetry.

a) Even; the graph is symmetric about the y-axis. b) Odd; the graph has origin symmetry.

- Test the equations \(f(-x)=f(x)\) and \(f(-x)=-f(x)\). - Simplify \((-x)^2\) and \((-x)^3\) carefully. - Compare the entire expression after substituting \(-x\).

1. For \(h\), \(h(-x)=\frac{(-x)^2-4}{(-x)^3}=\frac{x^2-4}{-x^3}=-h(x)\). Therefore, \(h\) is odd, and its graph has origin symmetry. 2. For \(k\), \(k(-x)=\sin((-x)^2)-x=\sin(x^2)-x\). This is neither \(k(x)=\sin(x^2)+x\) nor \(-k(x)=-\sin(x^2)-x\). Therefore, \(k\) is neither even nor odd.

a) Odd; the graph has origin symmetry. b) Neither even nor odd.

- Check the domain first. - Replace \(x\) with \(-x\). - Simplify the squared expressions before comparing.

1. Since \(x^2+5>0\) for every real \(x\), the domain is \(\mathbb{R}\), which is symmetric about \(0\). 2. Substitute \(-x\): \(f(-x)=(-x)^2\sqrt{(-x)^2+5}=x^2\sqrt{x^2+5}=f(x)\). 3. Therefore, \(f\) is even.

The function is even, so its graph is symmetric about the y-axis.

- Use the exponent to decide whether each power function is even or odd. - Determine whether changing the sign of \(x\) changes the function value. - For an even function, \(f(-x) = f(x)\). - For an odd function, \(f(-x) = -f(x)\).

1. The function \(f(x) = x^4\) is even, so \(f(-x) = f(x)\). Therefore, \(f(-0.8) = 0.4096\) and \(f(-1.5) = 5.0625\). 2. The function \(g(x) = x^7\) is odd, so \(g(-x) = -g(x)\). 3. Since \(g(0.6) = 0.0279936\), the input that gives \(-0.0279936\) is \(-0.6\). 4. Since \(g(1.2) = 3.5831808\), the input that gives \(-3.5831808\) is \(-1.2\).

a) \(f(-0.8) = 0.4096\) and \(f(-1.5) = 5.0625\) b) The missing inputs are \(-0.6\) and \(-1.2\), in that order.

- Use inputs equally far from the proposed vertical axis. - Substitute \(-5+h\) and \(-5-h\). - Compare the resulting squared expressions.

1. Symmetry about \(x=-5\) requires equal outputs at inputs equally far from \(-5\). 2. \(f(-5+h)=h^2-2\). 3. \(f(-5-h)=(-h)^2-2=h^2-2\). 4. Therefore, \(f(-5+h)=f(-5-h)\) for every real \(h\), proving symmetry about \(x=-5\).

Both \(f(-5+h)\) and \(f(-5-h)\) simplify to \(h^2-2\). Therefore, the graph is symmetric about \(x=-5\).

- Apply the definition of an even function to the formula for an arbitrary input \(x\). - Once evenness is established, think about how y-axis symmetry changes the sign of an x-coordinate. - A single value such as \(f(0)\) cannot establish an identity that must hold for every input.

1. \(f(-x)=0.25(-x)^4=0.25x^4=f(x)\) for every real \(x\), so \(f\) is even. 2. Even-function symmetry reflects points across the y-axis. Therefore, \((3,20.25)\) requires the point \((-3,20.25)\). 3. The condition \(f(0)=0\) alone does not prove oddness. For example, \(f(1)=0.25\) and \(f(-1)=0.25\), not \(-0.25\). Thus, \(f(-x)=-f(x)\) is not true for all \(x\).

a) \(f(-x)=f(x)\), so \(f\) is even. b) \((-3,20.25)\) c) Having \(f(0)=0\) is not sufficient for oddness; here \(f(-1)=f(1)=0.25\), so \(f(-1)\ne -f(1)\).

- Replace \(x\) by \(-x\) and compare the result with \(-f(x)\). - Origin symmetry changes the signs of both coordinates. - Distinguish a condition that is necessary at one input from a definition that must hold for every input.

1. \(f(-x)=0.25(-x)^3=-0.25x^3=-f(x)\) for every real \(x\), so \(f\) is odd. 2. Origin symmetry sends \((x,y)\) to \((-x,-y)\). Therefore, \((6,54)\) requires \((-6,-54)\). 3. Oddness requires \(f(-x)=-f(x)\) for every input in a symmetric domain. The single condition \(f(0)=0\) checks only one input and cannot establish that identity.

a) \(f(-x)=-f(x)\), so \(f\) is odd. b) \((-6,-54)\) c) The condition \(f(0)=0\) checks only one input; oddness requires \(f(-x)=-f(x)\) for every input in the domain.

- Test each rule for even or odd symmetry. - Use the y-intercept to distinguish the two even functions. - Match each algebraic feature to the graph.

1. For \(f\), \(f(-x)=-f(x)\), so its graph is symmetric about the origin. Only graph c has this symmetry, so \(f\) matches graph c. 2. For \(g\), \(g(-x)=g(x)\), so it is even. Also, \(g(0)=\cos(0)-1=0\). The even graph through the origin is graph a. 3. For \(h\), \(h(-x)=h(x)\), so it is even. Its y-intercept is \(h(0)=2\), identifying graph b.

Graph a: \(g(x)=\cos(0.5x)-1\); Graph b: \(h(x)=\frac{4}{x^2+2}\); Graph c: \(f(x)=\frac{4x}{x^2+2}\)

- Prove the parity identity for an arbitrary input before using the marked point. - For an odd function, what happens to both coordinates of a point under the graph symmetry? - Compare the point transformation for origin symmetry with a reflection across a single axis.

1. \(h(-x)=-0.2(-x)^3=0.2x^3=-h(x)\) for every real \(x\). Therefore, \(h\) is odd. 2. Origin symmetry maps \((x,y)\) to \((-x,-y)\). Thus, \(A(3,-5.4)\) corresponds to \(B(-3,5.4)\). 3. An x-axis reflection would keep the x-coordinate \(3\), while a y-axis reflection would keep the y-coordinate \(-5.4\). Here both coordinate signs change, which is the point transformation for origin symmetry.

a) \(h(-x)=-h(x)\), so \(h\) is odd. b) \(B=(-3,5.4)\) c) Both coordinate signs change, so the symmetry is about the origin.

- Identify the number whose cube is \(-8\). - Consider how many times a horizontal line can cross a strictly increasing cubic graph, and recall that an odd graph has origin symmetry. - A vertical compression uses a coefficient whose absolute value is between \(0\) and \(1\); test \(g(-x)\) to classify the symmetry.

1. Solve \(x^3=-8\): \(x=\sqrt[3]{-8}=-2\). 2. The function \(f(x)=x^3\) is strictly increasing and takes every real output value exactly once, so the line \(y=5\) intersects the graph once. Also, \(f(-x)=-f(x)\), so the graph is odd: if \((r, 5)\) is the intersection with \(y=5\), then \((-r, -5)\) is the corresponding intersection with \(y=-5\). 3. A vertical compression occurs when \(0<|a|<1\). One example is \(g(x)=0.5x^3\). Since \(g(-x)=-g(x)\), this function is also odd.

a) \(x=-2\) b) One intersection point. Odd symmetry gives a corresponding intersection with \(y=-5\) at the opposite x-coordinate. c) Sample answer: \(g(x)=0.5x^3\), which is odd.

- Check whether either function has restrictions on its input. - Determine whether each function can produce negative outputs. - Compare \(f(-x)\) with \(f(x)\), and do the same for \(g\).

1. Both functions are defined for every real input, so both domains are \((-\infty, \infty)\). 2. The cubic function takes every real value, so the range of \(f\) is \((-\infty, \infty)\). The sixth-power function is never negative, so the range of \(g\) is \([0, \infty)\). 3. Since \(f(-x)=-f(x)\), \(f\) is odd and its graph is symmetric about the origin. Since \(g(-x)=g(x)\), \(g\) is even and its graph is symmetric about the y-axis.

For \(f(x)=x^3\): domain \((-\infty, \infty)\), range \((-\infty, \infty)\), symmetric about the origin For \(g(x)=x^6\): domain \((-\infty, \infty)\), range \([0, \infty)\), symmetric about the y-axis

- Compare y-axis symmetry with origin symmetry. - Check whether each graph passes through the origin. - Identify the graph that approaches the coordinate axes as asymptotes.

1. Graph \(g\) is symmetric about the y-axis and has only nonnegative outputs, so it represents \(y=x^2\). 2. Graph \(h\) passes through the origin and is symmetric about the origin, with branches in Quadrants I and III. It represents \(y=x^3\). 3. Graph \(f\) has two separate branches in Quadrants I and III and approaches both coordinate axes without touching them. It represents \(y=\frac{1}{x}\).

\(f: y=\frac{1}{x}\); \(g: y=x^2\); \(h: y=x^3\)

- Use the relationship between origin symmetry and odd functions. - Determine what origin symmetry requires of the constant term. - Substitute the given point to find \(a\).

1. Symmetry about the origin requires \(f\) to be odd. Therefore, \(n\) must be odd and \(c=0\). 2. Since \((1,-2)\) lies on the graph, \(-2=a(1)^n\), so \(a=-2\). 3. The smallest odd integer greater than \(1\) is \(n=3\). Thus, \(f(x)=-2x^3\).

\(a=-2\), \(n=3\), and \(c=0\)

- Compare the outputs at opposite inputs. - Which parity of exponent gives y-axis symmetry? - Use the smallest positive even exponent, then substitute one point to find \(a\).

1. Because \(f(2)=16\), the coefficient \(a\) is nonzero. The two point conditions give \(a2^n=a(-2)^n\), so \(2^n=(-2)^n\). This equality holds only when \(n\) is even. 2. The smallest positive even integer is \(n=2\). 3. Substitute \((2,16)\) into \(f(x)=ax^2\): \(16=a(2^2)=4a\), so \(a=4\). 4. Therefore, \(f(x)=4x^2\).

a) Since \(a\ne0\) and \(a2^n=a(-2)^n\), \(n\) must be even. b) \(f(x)=4x^2\)

- Identify which function contains only even powers. - Identify which function contains only odd powers. - Compare those algebraic symmetry types with the visible graph symmetries.

1. The function \(f\) contains only even powers of \(x\), so it is even. Its graph is symmetric about the y-axis, which matches graph a. 2. The function \(g\) contains only odd powers of \(x\), so it is odd. Its graph is symmetric about the origin, which matches graph b. 3. The function \(h\) contains both even and odd powers, so it is neither even nor odd. Its graph has neither y-axis symmetry nor origin symmetry, which matches graph c.

\(a \to f\), \(b \to g\), and \(c \to h\)

- Identify the odd function first. - Distinguish the even functions by periodicity and end behavior. - Check what each function approaches for large positive \(x\).

1. Since \(a(-x)=-a(x)\), \(a\) is odd. The only graph symmetric about the origin is p, so p represents \(a\). 2. Both \(b\) and \(c\) are even. The cosine function oscillates between \(-1\) and \(1\), so the periodic graph q represents \(b\). 3. For \(c\), the exponent \(-0.2x^2\to-\infty\), so \(c(x)\to0\). The even graph that approaches the x-axis is r.

Graph p: \(a(x)=\frac{2x}{x^2+1}\); Graph q: \(b(x)=\cos(x)\); Graph r: \(c(x)=e^{-0.2x^2}\)

- Apply the odd-function definition to \(g\) before looking for other graph features. - Compare the outputs shown at opposite inputs \(-1\) and \(1\). - One violated required symmetry condition is enough to reject the proposed graph.

1. \(g(-x)=(-x)^3-3(-x)=-x^3+3x=-(x^3-3x)=-g(x)\). Therefore, \(g\) is odd. 2. From the displayed graph, \(k(-1)=4\) and \(k(1)=0\). 3. If \(k\) were odd, then \(k(-1)\) would have to equal \(-k(1)=0\). Since \(4\ne0\), the displayed graph is not odd. 4. The graph of \(g\) must be odd, but the displayed graph is not. Therefore, the displayed graph cannot represent \(g\).

a) \(g(-x)=-g(x)\), so \(g\) is odd. b) \(k(-1)=4\) and \(k(1)=0\), so \(k(-1)\ne-k(1)\); therefore, \(k\) is not odd. c) An odd function cannot have a graph that fails the odd-function test, so the displayed graph cannot be \(g\).

- Compare \(h(-x)\) with \(h(x)\). - Use the zero-product property on the factored expression. - Convert each zero into its corresponding x-intercept.

1. The graph is symmetric about the y-axis. Algebraically, \(h(-x) = 0.1(-x)^4 - 1.6(-x)^2 = 0.1x^4 - 1.6x^2 = h(x)\), so \(h\) is even. 2. The y-intercept is \((0, 0)\). The x-intercepts occur where \(h(x) = 0\). 3. To find and confirm the zeros, solve \(0.1x^2(x^2 - 16) = 0\). 4. The zero-product property gives \(x^2 = 0\) or \(x^2 - 16 = 0\). Therefore, \(x = 0\), \(x = -4\), or \(x = 4\), so the x-intercepts are \((-4, 0)\), \((0, 0)\), and \((4, 0)\).

a) The graph is symmetric about the y-axis. b) The x-intercepts are \((-4, 0)\), \((0, 0)\), and \((4, 0)\). The y-intercept is \((0, 0)\). c) Solving \(0.1x^2(x^2 - 16) = 0\) confirms the zeros \(x = -4\), \(x = 0\), and \(x = 4\).

- Substitute the first point into the general form to find \(a\). - Use the second point and the value of \(a\) to determine \(n\). - How does an even or odd exponent affect a function's symmetry?

1. Substitute \(P(1, -2)\): \(-2 = a(1)^n\), so \(a = -2\). 2. Substitute \(Q(2, -32)\): \(-32 = -2(2^n)\). 3. Divide by \(-2\): \(16 = 2^n\), so \(n = 4\). Therefore, \(f(x) = -2x^4\). 4. Because \(n=4\) is even, \(f(-x)=f(x)\). The graph is symmetric about the y-axis.

\(f(x) = -2x^4\). Because the exponent \(4\) is even, the graph is symmetric about the y-axis.

- Translate the stated symmetry into a condition on a power function's exponent. - Use the given point to determine what must be true about the coefficient. - Check whether those two conditions are compatible with decreasing behavior.

1. Symmetry about the origin requires \(n\) to be odd. 2. Since the graph passes through \((2,8)\), \(a\cdot2^n=8\), so \(a=\frac{8}{2^n}>0\). 3. A power function \(f(x)=ax^n\) with odd positive integer \(n\) and \(a>0\) is strictly increasing over all real numbers. 4. This contradicts the requirement that the function be strictly decreasing. Therefore, no such function exists.

No such function exists. Origin symmetry requires an odd exponent, and the point \((2,8)\) forces \(a>0\). That combination produces a strictly increasing function, not a decreasing one.

- Replace every \(x\) with \(-x\). - Distinguish the condition for one graph to be symmetric from the condition for two graphs to be reflections of each other. - Compare \(g(-x)\) with both \(g(x)\) and \(h(x)\).

1. Replace \(x\) with \(-x\): \(f(-x)=0.25(-x)^4-2(-x)^2+1=0.25x^4-2x^2+1=f(x)\). Therefore, \(f\) is even and its graph is symmetric about the y-axis. 2. Compute \(g(-x)=(-x)^3+2=-x^3+2=h(x)\). Thus, reflecting the graph of \(g\) across the y-axis produces the graph of \(h\). 3. For \(g\) to be symmetric about the y-axis, it would need to satisfy \(g(-x)=g(x)\). Instead, \(-x^3+2\ne x^3+2\) for nonzero \(x\), so \(g\) is not even.

a) \(f(-x)=f(x)\), so \(f\) is symmetric about the y-axis. b) \(g(-x)=h(x)\), so the graphs of \(g\) and \(h\) are reflections across the y-axis. c) \(g(-x)\ne g(x)\) for nonzero \(x\), so \(g\) is not symmetric about the y-axis.

- Test whether \(p(-x)=p(x)\) or \(p(-x)=-p(x)\). - Simplifying the products may make the powers easier to inspect. - One counterexample is enough to rule out a proposed identity.

1. For \(f\), simplify to \(f(x)=x(x^2-4)=x^3-4x\). Then \(f(-x)=-x^3+4x=-f(x)\), so \(f\) is odd. 2. For \(g\), \(g(-x)=((-x)^2-9)((-x)^2+1)=(x^2-9)(x^2+1)=g(x)\), so \(g\) is even. 3. For \(h\), neither identity holds. For example, \(h(2)=3\) and \(h(-2)=-9\), so \(h(-2)\ne h(2)\) and \(h(-2)\ne-h(2)\). Therefore, \(h\) is neither even nor odd.

a) Odd; symmetric about the origin b) Even; symmetric about the y-axis c) Neither even nor odd

- Replace every \(x\) with \(-x\) and factor out \(-1\). - Connect the identity \(g(-x)=-g(x)\) with odd functions. - Substitute each point to create a system of two equations.

1. Substitute \(-x\): \(g(-x)=a(-x)+\frac{b}{-x}=-ax-\frac{b}{x}=-\left(ax+\frac{b}{x}\right)=-g(x)\). 2. Therefore, \(g\) is odd, so its graph is symmetric about the origin. 3. The point \(P(1, 5)\) gives \(a+b=5\). The point \(Q(2, 8.5)\) gives \(2a+\frac{b}{2}=8.5\), or \(4a+b=17\). Subtracting the first equation gives \(3a=12\), so \(a=4\). Then \(b=1\).

1) \(g(-x)=-g(x)\) 2) The graph is symmetric about the origin. 3) \(a=4\) and \(b=1\)

- For polynomials, compare which powers of \(x\) remain after simplification. - For the rational function, test the effect of replacing \(x\) with \(-x\). - Expand the shifted square before classifying it. - A mixture of nonzero even- and odd-power terms usually makes a polynomial neither even nor odd.

1. Functions 1, 5, 8, 11, and 13 contain only even powers of \(x\) after simplification, so they satisfy \(f(-x)=f(x)\) and are even. 2. Functions 2, 3, 6, 9, and 12 satisfy \(f(-x)=-f(x)\), so they are odd. For function 3, replacing \(x\) with \(-x\) changes \(\frac{2}{x}\) to \(-\frac{2}{x}\). 3. Functions 4, 7, 10, 14, and 15 are neither. For example, \((x+3)^2=x^2+6x+9\) contains both even- and odd-power terms, so it is neither even nor odd.

Even: 1), 5), 8), 11), 13) Odd: 2), 3), 6), 9), 12) Neither: 4), 7), 10), 14), 15)

- Substitute \(-x\) into each product. - Replace each factor using its even or odd identity. - Track the signs when multiplying two negative factors.

1. For \(h\), \(h(-x)=f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x)=-h(x)\). Therefore, \(h\) is odd. 2. For \(k\), \(k(-x)=g_1(-x)g_2(-x)=(-g_1(x))(-g_2(x))=g_1(x)g_2(x)=k(x)\). Therefore, \(k\) is even.

\(h\) is odd and symmetric about the origin. \(k\) is even and symmetric about the y-axis.

- Substitute a strategic value into the odd-function identity. - Compare the values at opposite inputs. - Combine the equations for even and odd functions.

1. Since \(f\) is odd, \(f(-x)=-f(x)\). At \(x=0\), this gives \(f(0)=-f(0)\), so \(2f(0)=0\) and \(f(0)=0\). 2. If \(g\) were odd, \(g(-3)=-g(3)\). The given values would require \(4=-4\), which is impossible. 3. If \(f\) is both even and odd, then \(f(-x)=f(x)\) and \(f(-x)=-f(x)\). Thus, \(f(x)=-f(x)\), so \(2f(x)=0\) and \(f(x)=0\) for every \(x\) in the domain.

a) \(f(0)=-f(0)\), so \(f(0)=0\). b) Odd symmetry would require \(g(-3)=-g(3)\), but the given values are both \(4\). c) The two identities imply \(f(x)=-f(x)\), so \(f(x)=0\) for every \(x\) in the domain.

- Test a simple translated parabola at opposite x-values. - Apply the odd-function identity to \(g(-x)\). - Compare \(k(-x)\) with \(-k(x)\).

1. Choose \(f(x)=x^2\) and \(c=1\). Then \(h(x)=(x-1)^2\). Since \(h(1)=0\) but \(h(-1)=4\), \(h(-x)=h(x)\) does not hold in general. The translation moves the symmetry axis from \(x=0\) to \(x=1\). 2. Since \(g\) is odd, \(k(-x)=g(-x)+d=-g(x)+d\). However, \(-k(x)=-g(x)-d\). These expressions are equal only when \(d=0\), so for \(d\ne0\), \(k\) is not symmetric about the origin.

a) For example, \(f(x)=x^2\) is even, but \(h(x)=(x-1)^2\) is not even because \(h(1)=0\ne4=h(-1)\). b) No. For \(d\ne0\), \(k(-x)=-g(x)+d\ne-g(x)-d=-k(x)\).

- Recall whether sine and cosine are even or odd. - Track the signs in the numerator and denominator of the quotient. - For an odd polynomial, every nonzero term must have an odd exponent.

1. Since \(\cos(-x)=\cos(x)\) and \((-x)^2=x^2\), \(p(-x)=p(x)\). Thus, \(p\) is even. 2. Since \(\sin(-x)=-\sin(x)\) and \((-x)^3=-x^3\), \(q(-x)=\frac{-\sin(x)}{-x^3}=q(x)\). The domain is symmetric about \(0\), so \(q\) is even. 3. For \(f_a\) to be odd, its even-power term must vanish. Thus, \(a+1=0\), so \(a=-1\). Then \(f_{-1}(x)=x^3+x\), which is odd.

a) Even; symmetric about the y-axis b) Even; symmetric about the y-axis c) \(a=-1\)

- Use the identity \(g_k(-x)=-g_k(x)\). - The numerator is already even; determine when the denominator is odd. - Check that the resulting domain is symmetric about \(0\).

1. The numerator \(x^2+4\) is even. For the quotient to be odd, the linear denominator must be odd. 2. Rewrite the denominator as \(x+(5-k)\). A linear function of this form is odd only when its constant term is \(0\). 3. Set \(5-k=0\), which gives \(k=5\). 4. Verification: \(g_5(x)=\frac{x^2+4}{x}=x+\frac{4}{x}\), with domain \(x\ne0\). Both terms are odd, so \(g_5(-x)=-g_5(x)\). 5. Therefore, exactly one value of \(k\) works.

Exactly one value works: \(k=5\).

- Use only odd powers in the general cubic form. - Translate the zero and the point into equations. - Solve the resulting system for the coefficients.

1. An odd cubic polynomial has the form \(g(x)=ax^3+bx\). 2. The zero at \(x=2\) gives \(8a+2b=0\), or \(4a+b=0\). 3. The point \((1, 6)\) gives \(a+b=6\). 4. Subtracting the second equation from the first relation gives \(3a=-6\), so \(a=-2\). Then \(b=8\). 5. Therefore, \(g(x)=-2x^3+8x\).

\(g(x)=-2x^3+8x\)

- Identify the even-power and odd-power terms. - An odd polynomial cannot have a nonzero even-power term. - An even polynomial cannot have a nonzero cubic term.

1. For \(f\) to be odd, the coefficient of its even-power term must be \(0\). Thus, \(a+2=0\), so \(a=-2\). 2. The coefficient \(b\) multiplies the odd-power term \(x\), so it may be any real number. Therefore, all pairs \((-2, b)\), where \(b\in\mathbb{R}\), make \(f\) odd. 3. For \(f\) to be even, every odd-power coefficient would need to be \(0\). The coefficient of \(x^3\) is fixed at \(1\), so no values of \(a\) and \(b\) can make \(f\) even.

Odd: all pairs \((-2, b)\) with \(b\in\mathbb{R}\) Even: no ordered pairs

- Apply \(f(-x)=f(x)\) at a zero. - Count nonzero zeros in opposite pairs. - Build an even polynomial from factors for \(-1\), \(0\), and \(1\).

1. Since \(f\) is even, \(f(-x)=f(x)\). If \(f(x_0)=0\), then \(f(-x_0)=f(x_0)=0\), so nonzero zeros occur in opposite pairs. 2. Three is odd, while the nonzero zeros occur in pairs. Therefore, the unpaired zero must be \(x=0\), which is its own opposite. 3. Choose the distinct zeros \(-1, 0, 1\). The even polynomial \(f(x)=x^2(x-1)(x+1)=x^4-x^2\) has exactly those three distinct real zeros.

1) If \(f(x_0)=0\), then \(f(-x_0)=0\). 2) The function must have the zero \(x=0\). 3) One possible rule is \(f(x)=x^4-x^2\).

- Substitute \(x=0\) into the odd-function identity. - Determine what symmetry does to a nonzero x-intercept. - Count the origin together with opposite pairs.

1. Odd symmetry gives \(g(-x)=-g(x)\). Setting \(x=0\) yields \(g(0)=-g(0)\), so \(2g(0)=0\) and \(g(0)=0\). 2. Every nonzero zero \(x_0\) has a matching zero \(-x_0\), while \(0\) is always a zero. Thus, the number of distinct real zeros is \(1\) plus an even number, so it cannot equal \(2\). 3. Since \(4\) is a zero, \(-4\) is also a zero. The function must also have the zero \(0\).

1) \(g(0)=0\) 2) Its distinct real zeros consist of \(0\) and zero or more opposite pairs, so the total cannot be \(2\). 3) The additional required zeros are \(-4\) and \(0\).

- Set the coefficients of all even-power terms equal to \(0\). - Substitute the coordinates of the given point. - Solve the remaining linear equation.

1. For \(f\) to be odd, the coefficients of the even-power terms must be \(0\). Thus, \(2a=0\) and \(c=0\), so \(a=0\) and \(c=0\). 2. The function becomes \(f(x)=(b-1)x^3+4x\). 3. Since \((1, 6)\) lies on the graph, \(6=(b-1)+4=b+3\). 4. Therefore, \(b=3\).

\(a=0\), \(b=3\), and \(c=0\)

- Replace \(x\) with \(-x\) in each rule. - Use \(|-x|=|x|\). - Compare the result with both the original function and its negative.

1. For \(f\), \(f(-x)=\frac{(-x)^2+3}{(-x)^4}=f(x)\). Its domain excludes \(0\) but is symmetric about \(0\), so \(f\) is even. 2. Since \(|-x|=|x|\), \(g(-x)=(-x)|-x|=-x|x|=-g(x)\). Thus, \(g\) is odd. 3. \(h(-x)=e^{-x}+e^x=h(x)\), so \(h\) is even. 4. \(k(-x)=-x^3+x+1\). This is neither \(k(x)=x^3-x+1\) nor \(-k(x)=-x^3+x-1\), so \(k\) is neither even nor odd.

a) Even b) Odd c) Even d) Neither even nor odd

- Compute \(f(-x)\) carefully. - Compare it first with \(f(x)\), then with \(-f(x)\). - A symmetry identity must hold for every value in the domain.

1. Substitute \(-x\): \(f(-x)=\frac{x^2-3x}{x^2+4}\). 2. This is not equal to \(f(x)=\frac{x^2+3x}{x^2+4}\) for all \(x\), so \(f\) is not even. 3. Also, \(-f(x)=\frac{-x^2-3x}{x^2+4}\), which is not equal to \(f(-x)\) for all \(x\). Therefore, \(f\) is not odd. 4. The graph has neither symmetry.

The graph is neither symmetric about the y-axis nor symmetric about the origin.

- Consider even and odd exponents separately. - Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Use the shapes of familiar graphs such as \(y=x^2\) and \(y=x^3\). - Determine which output values are possible in each case.

1. If \(n\) is even, then \(f(-x)=0.4(-x)^n=0.4x^n=f(x)\), so \(f\) is even and its graph is symmetric about the y-axis. The function decreases on \((-\infty, 0]\) and increases on \([0, \infty)\). Its range is \([0, \infty)\). 2. If \(n\) is odd, then \(f(-x)=0.4(-x)^n=-0.4x^n=-f(x)\), so \(f\) is odd and its graph has rotational symmetry about the origin. The function increases on \((-\infty, \infty)\), and its range is \((-\infty, \infty)\).

Even \(n\): even; symmetric about the y-axis; decreasing on \((-\infty, 0]\); increasing on \([0, \infty)\); range \([0, \infty)\). Odd \(n\): odd; rotational symmetry about the origin; increasing on \((-\infty, \infty)\); range \((-\infty, \infty)\).

- Compare the overall behavior of \(x^2\) and \(x^3\). - Think about how multiplying all function values by a negative number changes a graph. - Use the definitions of even and odd functions to determine whether the sign of \(a\) affects symmetry.

1. For odd \(n\), \(x^n\) is strictly increasing on all real numbers. Multiplying by a negative coefficient reverses the order of the outputs, so \(ax^n\) is strictly decreasing exactly when \(n\) is odd and \(a<0\). For even \(n\), the function changes direction at \(x=0\), so it cannot be strictly decreasing over its entire domain. 2. Replacing \(a\) with \(-a\) reflects the graph across the x-axis. The parity of \(n\), not the sign of \(a\), determines the symmetry. When \(n\) is even, both graphs are symmetric about the y-axis. When \(n\) is odd, both graphs have rotational symmetry about the origin.

a) \(n\) must be odd and \(a<0\). b) No. Replacing \(a\) with \(-a\) reflects the graph across the x-axis but preserves y-axis symmetry for even \(n\) and origin symmetry for odd \(n\).

- The opposite-input numerical values in part a are evidence, not a general proof. - In part b, substitute \(-z\) into each entire function and simplify. - Use whether the result equals the original function or its negative to classify the function. - Apply that classification directly in part c instead of recomputing the powers.

1. a) \(A(3)=81-45=36\) and \(A(-3)=81-45=36\). Also, \(B(3)=27-3=24\) and \(B(-3)=-27+3=-24\). 2. b) \(A(-z)=(-z)^4-5(-z)^2=z^4-5z^2=A(z)\), so \(A\) is even. 3. \(B(-z)=(-z)^3-(-z)=-z^3+z=-(z^3-z)=-B(z)\), so \(B\) is odd. 4. c) Since \(A\) is even, \(A(-2)=A(2)=-4\). Since \(B\) is odd, \(B(-2)=-B(2)=-6\).

a) \(A(3)=A(-3)=36\); \(B(3)=24\), \(B(-3)=-24\) b) \(A(-z)=A(z)\), so \(A\) is even; \(B(-z)=-B(z)\), so \(B\) is odd. c) \(A(-2)=-4\) and \(B(-2)=-6\)

- Recall which powers can appear in an even polynomial function. - Determine which coefficient must equal zero. - Substitute \(-x\) for \(x\) to verify the even-function condition.

1. A polynomial is even only when the coefficients of all odd-power terms are zero. 2. The coefficient of \(x^3\) must be zero, so \(a + 3 = 0\). Therefore, \(a = -3\). 3. Substituting \(a = -3\) gives \(f(x) = 2x^4 - 5x^2\). 4. Then \(f(-x) = 2(-x)^4 - 5(-x)^2 = 2x^4 - 5x^2 = f(x)\). Therefore, the graph is symmetric about the y-axis.

\(a = -3\)

- Recall the equation that defines an odd function. - Track the sign of each term after substituting \(-x\) for \(x\). - Determine how adding the constant \(5\) affects the equation \(h(-x) = -h(x)\).

1. Substitute \(-x\): \(g(-x) = \frac{1}{2}(-x)^5 - 4(-x)^3 + (-x) = -\frac{1}{2}x^5 + 4x^3 - x\). 2. This expression equals \(-g(x)\), so \(g(-x) = -g(x)\). Therefore, \(g\) is odd and its graph is symmetric about the origin. 3. For \(h(x) = g(x) + 5\), \(h(-x) = g(-x) + 5 = -g(x) + 5\). 4. However, \(-h(x) = -g(x) - 5\). Since \(h(-x) \ne -h(x)\), \(h\) is not odd and its graph is not symmetric about the origin.

1. Yes. \(g(-x) = -g(x)\), so \(g\) is symmetric about the origin. 2. No. \(h(-x) = -g(x) + 5\), while \(-h(x) = -g(x) - 5\).

- Identify which powers are even and which are odd. - Determine which coefficients can be changed by choosing \(a\), \(b\), and \(c\). - A term disappears only when its coefficient is zero.

1. For symmetry about the origin, \(g\) must be odd. Therefore, the coefficients of the even-power terms must be zero. 2. Set \(a - 1 = 0\), \(b + 3 = 0\), and \(c = 0\). This gives \(a = 1\), \(b = -3\), and \(c = 0\). 3. The resulting function is \(g(x) = 2x^3\), which is odd. 4. For symmetry about the y-axis, \(g\) would have to be even, so every odd-power term would need coefficient zero. 5. The coefficient of \(x^3\) is fixed at \(2\), so that term cannot be eliminated. Therefore, no parameter values make \(g\) even.

a) \(a = 1\), \(b = -3\), and \(c = 0\) b) No such values exist because the fixed term \(2x^3\) prevents the function from being even.

- Determine which exponents make \(ax^n + c\) an even function. - Evaluate \(f(0)\) to find the y-intercept. - Consider how the sign of \(a\) affects the end behavior when \(n\) is even.

1. Symmetry about the y-axis requires \(f\) to be even, so \(n\) must be a positive even integer. 2. The y-intercept is determined by \(f(0) = c\). Therefore, \(c = 4\). 3. For a positive even exponent, \(x^n \to \infty\) as \(x \to \infty\). To make \(ax^n + c \to -\infty\), the leading coefficient must satisfy \(a < 0\).

\(n \in \{2, 4, 6, \ldots\}\), \(c = 4\), and \(a < 0\)

- Recall the relationship between origin symmetry and odd functions. - Substitute each point into the function equation. - Solve the resulting system of two linear equations.

1. Symmetry about the origin means that \(f\) is odd, so only odd powers of \(x\) may have nonzero coefficients. Therefore, \(b = 0\) and \(d = 0\). 2. The function becomes \(f(x) = ax^3 + cx\). Substituting \(P(1, 5)\) gives \(a + c = 5\). 3. Substituting \(Q(2, 22)\) gives \(8a + 2c = 22\), or \(4a + c = 11\). 4. Subtracting \(a + c = 5\) from \(4a + c = 11\) gives \(3a = 6\), so \(a = 2\). 5. Then \(2 + c = 5\), so \(c = 3\).

a) \(b = 0\) and \(d = 0\), because an odd polynomial contains only odd powers of \(x\). b) \(a = 2\) and \(c = 3\)

- Use the relationship between origin symmetry and odd functions. - Identify which term determines the end behavior. - Consider the cases \(n < 3\), \(n = 3\), and \(n > 3\). - When \(n = 3\), combine the like terms.

1. Symmetry about the origin requires \(f\) to be odd, so \(n\) must be odd. 2. The leading term must have degree \(3\) and leading coefficient \(-4\). 3. If \(n = 1\), then \(f(x) = 2x + cx^3\). The leading term is \(cx^3\), so \(c = -4\). 4. If \(n = 3\), then \(f(x) = (2 + c)x^3\). Setting \(2 + c = -4\) gives \(c = -6\). 5. If \(n \ge 5\) is odd, then \(2x^n\) has degree greater than \(3\) and becomes the leading term, so the required end behavior is impossible.

\((n, c) = (1, -4)\) or \((n, c) = (3, -6)\)

- Use the relationship between y-axis symmetry and even functions. - Identify which term determines the end behavior. - Consider the cases \(m < 4\), \(m = 4\), and \(m > 4\). - Remember that \(0\) is an even exponent.

1. Symmetry about the y-axis requires \(h\) to be even, so \(m\) must be even. 2. If \(m < 4\), then \(m = 0\) or \(m = 2\). The term \(ax^4\) is the leading term, so \(a = 5\). 3. If \(m = 4\), then \(h(x) = (a + 5)x^4\). Setting \(a + 5 = 5\) gives \(a = 0\). 4. If \(m > 4\), then the term \(5x^m\) has degree greater than \(4\), so the leading term cannot be \(5x^4\).

\((a, m) = (5, 0)\), \((5, 2)\), or \((0, 4)\)

- Use the condition \(f(-x) = -f(x)\) for origin symmetry. - Simplify powers with odd exponents. - Interpret adding \(3\) as a vertical translation. - Track how a vertical translation changes the center of symmetry.

1. Substitute \(-x\) for \(x\): \(p(-x) = \frac{1}{4}(-x)^5 - 2(-x)^3\). 2. Because \(5\) and \(3\) are odd, \(p(-x) = -\frac{1}{4}x^5 + 2x^3 = -p(x)\). Therefore, the graph of \(p\) is symmetric about the origin. 3. The function \(q(x) = p(x) + 3\) is the graph of \(p\) shifted up \(3\) units. 4. Since \(q(-x) - 3 = p(-x) = -p(x) = -(q(x) - 3)\), the graph of \(q\) is symmetric about the point \((0, 3)\). It is not symmetric about the origin or the y-axis.

a) \(p(-x) = -p(x)\), so the graph is symmetric about the origin. b) The graph of \(q\) is symmetric about \((0, 3)\).

- Begin with \(f(-x) = -f(x)\). - Substitute \(-x\) into each new function. - Compare each result with the original function and its opposite. - Recall that squaring a negative expression gives the same result as squaring the positive expression.

1. Because \(f\) is odd, \(f(-x) = -f(x)\). 2. For \(g\), \(g(-x) = (-x)f(-x) = (-x)(-f(x)) = xf(x) = g(x)\). Therefore, \(g\) is even. 3. For \(h\), \(h(-x) = (f(-x))^2 = (-f(x))^2 = (f(x))^2 = h(x)\). Therefore, \(h\) is even. 4. For \(k\), \(k(-x) = \frac{1}{2}f(-x) = -\frac{1}{2}f(x) = -k(x)\). Therefore, \(k\) is odd.

a) Even; the graph is symmetric about the y-axis. b) Even; the graph is symmetric about the y-axis. c) Odd; the graph is symmetric about the origin.

- Use \(f(-x) = f(x)\) for an even function. - Substitute \(-x\) into each function. - Compare the result with both the original function and its opposite. - Pay special attention to the fact that \(-0 = 0\).

1. Since \(f\) is even, \(f(-x) = f(x)\). Therefore, \(g(-x) = (-x)f(-x) = -xf(x) = -g(x)\), so \(g\) is odd. 2. For \(h(x) = 5\), \(h(-x) = 5 = h(x)\), so \(h\) is even. However, \(h(-x) = 5 \ne -5 = -h(x)\), so \(h\) is not odd. 3. For \(k(x) = 0\), \(k(-x) = 0 = k(x)\) and \(k(-x) = 0 = -k(x)\). Therefore, the zero function is both even and odd.

a) \(g\) is odd; its graph is symmetric about the origin. b) \(h(x) = 5\) is even but not odd. \(k(x) = 0\) is both even and odd.

- Use the defining equation for an odd function. - Substitute \(-x\) into the sum and factor out \(-1\). - A vertical shift changes the function by adding a constant. - Compare \(h(-x)\) directly with \(-h(x)\).

1. Since \(p\) and \(q\) are odd, \(p(-x) = -p(x)\) and \(q(-x) = -q(x)\). 2. Then \(s(-x) = p(-x) + q(-x) = -p(x) - q(x) = -(p(x) + q(x)) = -s(x)\). Therefore, \(s\) is odd. 3. The shifted function is \(h(x) = x^3 - 4x + 3\). 4. Substituting \(-x\) gives \(h(-x) = -x^3 + 4x + 3\), while \(-h(x) = -x^3 + 4x - 3\). 5. Since \(h(-x) \ne -h(x)\), the shifted function is not odd.

a) \(s\) is odd because \(s(-x) = -s(x)\). b) No. The function \(h(x) = x^3 - 4x + 3\) is not odd.

- Recenter the input first; do not test \(g\) itself for oddness about the origin. - Apply the definition of an odd function to the new function \(h\). - Translate the identity for \(h(t)\) and \(h(-t)\) back into statements about \(g\).

1. Substitute \(u+2\) into \(g\): \(h(u)=g(u+2)=u^3\). 2. Then \(h(-u)=(-u)^3=-u^3=-h(u)\), so \(h\) is odd. 3. Because \(h\) is odd, \(h(t)=-h(-t)\). By the definition of \(h\), this is \(g(2+t)=-g(2-t)\). 4. Inputs equally far to the left and right of \(x=2\) have opposite outputs, so the graph of \(g\) has point symmetry about \((2, 0)\).

a) \(h(u)=u^3\) b) \(h(-u)=-h(u)\), so \(h\) is odd. c) \(g(2+t)=-g(2-t)\) for every real \(t\); therefore, the graph of \(g\) has point symmetry about \((2, 0)\).

- Substitute \(-x\) for every occurrence of \(x\). - Track the signs of even and odd powers. - Compare each result with both the original function and its opposite. - Treat a constant as a term with exponent \(0\).

1. \(f(-x) = 4(-x)^6 - (-x)^2 + 3 = 4x^6 - x^2 + 3 = f(x)\). Therefore, \(f\) is even. 2. \(g(-x) = -2(-x)^3 - 5(-x) = 2x^3 + 5x = -g(x)\). Therefore, \(g\) is odd. 3. \(h(-x) = (-x)^4 + (-x) = x^4 - x\). This is neither \(h(x) = x^4 + x\) nor \(-h(x) = -x^4 - x\). Therefore, \(h\) is neither even nor odd.

1. Even; the graph is symmetric about the y-axis. 2. Odd; the graph is symmetric about the origin. 3. Neither even nor odd.

- Identify the terms with odd exponents. - An even polynomial must have coefficient \(0\) on every odd-power term. - Check whether one value of \(k\) satisfies both odd-power conditions. - Decide whether the coefficient of an even-power term affects evenness.

1. An even polynomial has no nonzero odd-power terms. 2. Set the coefficient of \(x^3\) equal to zero: \(k - 5 = 0\), so \(k = 5\). 3. With \(k = 5\), the coefficient of \(x\) is \(2k - 10 = 2(5) - 10 = 0\). 4. The term \(mx^2\) has an even exponent, so \(m\) may be any real number.

\(k = 5\) and \(m\) may be any real number.

- Expand the product completely. - Identify the even-power term. - An odd polynomial cannot have a nonzero even-power term. - Substitute \(-x\) into the resulting function to verify the definition.

1. Expand the expression: \(g(x) = x^3 + (a - 6)x^2 - 6ax\). 2. For \(g\) to be odd, the coefficient of the even-power term \(x^2\) must be zero. Thus, \(a - 6 = 0\), so \(a = 6\). 3. With \(a = 6\), \(g(x) = x^3 - 36x\). 4. Then \(g(-x) = (-x)^3 - 36(-x) = -x^3 + 36x = -(x^3 - 36x) = -g(x)\). Therefore, \(g\) is odd.

a) \(a = 6\) b) For \(a = 6\), \(g(-x) = -g(x)\), so the graph is symmetric about the origin.

- An odd polynomial may contain only odd powers of \(x\). - To disprove a statement, one valid counterexample is enough. - Look for a polynomial from which you can factor out \(x\). - A cubic polynomial can have as many as three real zeros.

1. Consider \(f(x) = x^3 - x\). Since \(f(-x) = -f(x)\), the function is odd. 2. Factor the polynomial: \(f(x) = x(x^2 - 1) = x(x - 1)(x + 1)\). 3. By the zero-product property, the real zeros are \(x = -1\), \(x = 0\), and \(x = 1\). 4. This odd polynomial has three real zeros, so the statement is false.

One counterexample is \(f(x) = x^3 - x\). Its real zeros are \(-1\), \(0\), and \(1\), so it does not have exactly one real zero.

- Track the sign of a negative number raised to the third power. - Compare the outputs of the two functions for the same input. - Use the tests \(g(-x)=g(x)\) for even functions and \(g(-x)=-g(x)\) for odd functions. - What does a negative sign in front of the entire function do to its graph?

1. For \(f(x)=x^3\), the values are \(-8,-1,0,1,8\) for \(x=-2,-1,0,1,2\), respectively. 2. For \(g(x)=-x^3\), the values are \(8,1,0,-1,-8\). 3. Multiplying every output of \(f\) by \(-1\) reflects its graph across the x-axis, producing the graph of \(g\). Because \(f\) is odd, reflecting across the y-axis also gives the same graph. 4. \(g(-x)=-(-x)^3=x^3\), while \(-g(x)=-(-x^3)=x^3\). Therefore, \(g(-x)=-g(x)\), so \(g\) is odd.

a) \(f(x): -8,-1,0,1,8\); \(g(x): 8,1,0,-1,-8\) b) Reflect the graph of \(f\) across the x-axis. For this function, reflection across the y-axis gives the same result. c) \(g\) is odd because \(g(-x)=-g(x)\).

- Use the given point to solve for the unknown coefficient. - Evaluate the square before multiplying by the coefficient. - Compare \(k(x)\) directly with \(h(x)\) by writing \(k(x)=ch(x)\). - A negative output scaling reflects a graph across the x-axis.

1. Substitute \((2,4)\): \(4=a(2^2)=4a\), so \(a=1\). 2. Then \(h(-3)=(-3)^2=9\). 3. Substitute \((1,-2)\) into \(k(x)=bx^2\): \(-2=b(1)^2\), so \(b=-2\). 4. Since \(k(x)=-2x^2=-2h(x)\), reflect the graph of \(h\) across the x-axis and stretch it vertically by a factor of \(2\). Except for the origin, the graph of \(k\) lies in Quadrants III and IV.

a) \(a=1\) b) \(h(-3)=9\) c) \(k(x)=-2h(x)\). Reflect the graph of \(h\) across the x-axis and stretch it vertically by a factor of \(2\). Except for the origin, the graph lies in Quadrants III and IV.

- Use the definitions \(f(-x)=f(x)\) for even functions and \(f(-x)=-f(x)\) for odd functions. - Substitute \(-x\) into the inner function first. - For part c, try using an outer function that reverses a horizontal shift.

1. If \(v\) is odd, then \(v(-x)=-v(x)\). If \(u\) is even, then \(u(-v(x))=u(v(x))\). Therefore, \(f(-x)=u(v(-x))=u(-v(x))=u(v(x))=f(x)\), so \(f\) is even. 2. If both functions are odd, then \(f(-x)=u(v(-x))=u(-v(x))=-u(v(x))=-f(x)\). Therefore, \(f\) is odd. 3. For example, let \(v(x)=x+1\), which is neither even nor odd, and let \(u(x)=(x-1)^2\). Then \(f(x)=u(v(x))=((x+1)-1)^2=x^2\), which is even.

a) \(f\) is even. b) \(f\) is odd. c) One example is \(v(x)=x+1\) and \(u(x)=(x-1)^2\), which give \(f(x)=x^2\).

- Replace \(x\) with \(-x\) and simplify. - Consider the even and odd cases for \(g\) separately in part b. - Use the commutative property of multiplication in part c.

1. The inner function \(x^4+1\) is even because \((-x)^4+1=x^4+1\). Therefore, \(h(-x)=f((-x)^4+1)=f(x^4+1)=h(x)\). 2. If \(g\) is even, then \(k(-x)=(g(-x))^2=(g(x))^2=k(x)\). If \(g\) is odd, then \(k(-x)=(-g(x))^2=(g(x))^2=k(x)\). Thus \(k\) is even in either case. 3. \(s(-x)=g(-x)g(-(-x))=g(-x)g(x)=g(x)g(-x)=s(x)\). Therefore, \(s\) is even.

a) \(h(-x)=h(x)\), so \(h\) is even. b) Squaring removes the possible negative sign, so \(k(-x)=k(x)\). c) \(s(-x)=g(-x)g(x)=s(x)\), so \(s\) is even.

- Test whether \(f_t(-x)=f_t(x)\). - Substitute the given point into the function. - Notice the symmetry between the positive and negative exponents. - Check whether more than one parameter value works.

1. Evaluate the function at \(-x\): \(f_t(-x)=e^{-tx}+e^{tx}=f_t(x)\). Therefore, each function is even and its graph is symmetric about the y-axis. 2. The point condition gives \(e^{2t}+e^{-2t}=e^4+e^{-4}\). 3. Let \(u=e^{2t}>0\). Then \(u+\frac{1}{u}=e^4+e^{-4}\), whose positive solutions are \(u=e^4\) and \(u=e^{-4}\). 4. Thus, \(2t=4\) or \(2t=-4\), giving \(t=2\) or \(t=-2\).

a) \(f_t(-x)=f_t(x)\), so each graph is symmetric about the y-axis. b) \(t=2\) or \(t=-2\)

- Determine the domain before testing symmetry. - Compare the effects of \(-x\) on the fourth and third powers. - Check whether the result equals \(g(x)\) or \(-g(x)\).

1. The denominator requires \(x\ne0\), while \(x^4+1>0\) for every real \(x\). Thus, the domain is \(\mathbb{R}\setminus\{0\}\), which is symmetric about \(0\). 2. Substitute \(-x\): \(g(-x)=\frac{\sqrt{(-x)^4+1}}{(-x)^3}=\frac{\sqrt{x^4+1}}{-x^3}=-g(x)\). 3. Therefore, \(g\) is odd.

The function is odd, so its graph is symmetric about the origin.

- Compare the function values at \(0.5\) and \(-0.5\). - Use the point \((2, 1024)\) to solve for the exponent. - Decide whether the exponent is even or odd. - Use symmetry to complete the remaining entries.

1. From \(h(2) = 1024\), solve \(2^n = 1024\). Since \(1024 = 2^{10}\), \(n = 10\). 2. The table shows \(h(0.5) = h(-0.5)\). Since \(n = 10\) is even, \(h(-x) = h(x)\), so the graph is symmetric about the y-axis. 3. By symmetry, \(h(-2) = h(2) = 1024\). 4. Since \(3^{10} = 59{,}049\), the missing positive input is \(3\). 5. By symmetry, \(h(-3) = h(3) = 59{,}049\).

a) \(n = 10\) b) The graph is symmetric about the y-axis. c) The missing entries, from top to bottom, are \(1024\), \(3\), and \(59{,}049\).

- A leading odd power does not by itself make an entire polynomial odd. - Use a necessary property of odd functions to test \(g\) before recentering it. - Simplify the translated function \(h\) and then apply the odd-function definition to \(h\).

1. An odd function must satisfy \(g(-x)=-g(x)\). Here \(g(0)=5\ne0\), so \(g\) cannot be odd. 2. Substitute \(u-1\): \(h(u)=g(u-1)-4=((u-1)+1)^3+4-4=u^3\). 3. Since \(h(-u)=(-u)^3=-u^3=-h(u)\), the recentered function \(h\) is odd. 4. The transformation \(h(u)=g(u-1)-4\) moves the point \((-1, 4)\) of the original coordinate system to the origin. Since the recentered graph is odd, the original graph has point symmetry about \((-1, 4)\), but it is not symmetric about the origin and therefore is not itself an odd function.

a) \(g\) is not odd; for example, \(g(0)=5\ne0\). b) \(h(u)=u^3\), so \(h\) is odd. c) The graph of \(g\) is a translated odd-function graph with center \((-1, 4)\), but \(g\) itself is not odd because its symmetry center is not the origin.

- Eliminate the even-power term. - Factor out \(x\) after substituting the parameter value. - Use the difference of squares.

1. For the polynomial to be odd, the coefficient of the even-power term must be \(0\): \(k-5=0\), so \(k=5\). 2. Then \(f(x)=x^3-36x\), which is odd and has a graph symmetric about the origin. 3. Factor the equation \(x^3-36x=0\): \(x(x^2-36)=x(x-6)(x+6)=0\). 4. The real zeros are \(-6\), \(0\), and \(6\).

\(k=5\); the graph is symmetric about the origin; the real zeros are \(x=-6, 0, 6\).

- Eliminate the odd-power term. - Set the resulting polynomial equal to \(0\). - Remember that an even power can produce two real roots.

1. For the polynomial to be even, the coefficient of the linear term must be \(0\): \(a+1=0\), so \(a=-1\). 2. Then \(g(x)=x^4-81\), which is even and has a graph symmetric about the y-axis. 3. Set \(x^4-81=0\). Then \(x^4=81\), so the real solutions are \(x=3\) and \(x=-3\).

\(a=-1\); the graph is symmetric about the y-axis; the real zeros are \(x=-3\) and \(x=3\).

- Compare \(p(-x)\) with \(p(x)\) and \(-p(x)\). - A polynomial with only even powers is even; one with only odd powers is odd. - Use the leading term to determine the end behavior.

1. The function \(f\) contains only even powers of \(x\), so \(f(-x)=f(x)\). Its graph is symmetric about the y-axis. Its leading term is \(-x^4\), so both ends fall toward \(-\infty\). 2. The function \(g\) contains only odd powers of \(x\), so \(g(-x)=-g(x)\). Its graph is symmetric about the origin. Its leading term is \(0.5x^5\), so the left end falls and the right end rises. 3. The function \(h\) contains both even and odd powers, so it is neither even nor odd. Its leading term is \(-x^3\), so the left end rises and the right end falls.

a) Symmetric about the y-axis; \(\lim_{x\to\infty}f(x)=-\infty\) and \(\lim_{x\to-\infty}f(x)=-\infty\). b) Symmetric about the origin; \(\lim_{x\to\infty}g(x)=\infty\) and \(\lim_{x\to-\infty}g(x)=-\infty\). c) Neither; \(\lim_{x\to\infty}h(x)=-\infty\) and \(\lim_{x\to-\infty}h(x)=\infty\).

- Try factoring by grouping so that a common binomial factor appears. - Read each zero and its multiplicity from the fully factored form. - Compare the graph behavior associated with odd and even multiplicities.

1. Factor by grouping: \(g_k(x)=x^2(x+k)-k^2(x+k)=(x+k)(x^2-k^2)=(x-k)(x+k)^2\). 2. Thus \(x=k\) is a zero of multiplicity \(1\), and \(x=-k\) is a zero of multiplicity \(2\). 3. A zero of odd multiplicity changes the sign of the polynomial, so the graph crosses the x-axis at \(x=k\). 4. A zero of even multiplicity does not change the sign of the polynomial, so the graph touches the x-axis and turns around at \(x=-k\).

a) \(g_k(x)=(x-k)(x+k)^2\). The zeros are \(x=k\) with multiplicity \(1\) and \(x=-k\) with multiplicity \(2\). b) The graph crosses at \(x=k\) and touches the x-axis and turns around at \(x=-k\).

- Use symmetry to identify whether only even or only odd powers should appear. - Read the zeros from each graph and write a factored form. - Crossing x-intercepts correspond to simple factors. - Substitute the given point to determine the scale factor. - Expand the factored expression into the requested standard form.

1. The zeros of \(f\) are \(x = -2\), \(x = -1\), \(x = 1\), and \(x = 2\). Y-axis symmetry pairs the zeros, so \(f(x) = A(x^2 - 1)(x^2 - 4)\). 2. Substitute \(P(0, 2)\): \(2 = A(-1)(-4) = 4A\), so \(A = 0.5\). Expanding gives \(f(x) = 0.5x^4 - 2.5x^2 + 2\). 3. The zeros of \(g\) are \(x = -3\), \(x = 0\), and \(x = 3\). Origin symmetry is consistent with \(g(x) = Bx(x^2 - 9)\). 4. Substitute \(Q(2, -2)\): \(-2 = B \cdot 2(4 - 9) = -10B\), so \(B = 0.2\). Expanding gives \(g(x) = 0.2x^3 - 1.8x\).

\(f(x) = 0.5x^4 - 2.5x^2 + 2\) \(g(x) = 0.2x^3 - 1.8x\)

- Rewrite the quadratic using \(x-2\), or substitute both expressions directly. - Compare the even powers of \(a\). - Interpret \(2+a\) and \(2-a\) as inputs equally far from the proposed axis.

1. Rewrite the quadratic as \(f(x)=(x-2)^2+3\). 2. Substitute \(x=2+a\): \(f(2+a)=a^2+3\). 3. Substitute \(x=2-a\): \(f(2-a)=(-a)^2+3=a^2+3\). 4. Therefore, \(f(2+a)=f(2-a)\) for every real \(a\). 5. The inputs \(2+a\) and \(2-a\) are equally far from \(2\). Equal outputs at every such pair mean the graph is symmetric about \(x=2\).

a) \(f(2+a)=f(2-a)=a^2+3\) b) Points equally far to the left and right of \(x=2\) have equal y-values, so the graph is symmetric about \(x=2\).

- Compute \(f(-x)\) for each function. - Compare \(f(-x)\) with both \(f(x)\) and \(-f(x)\). - Use one additional feature, such as an intercept or end behavior, to confirm each match.

1. For \(f_1\), \(f_1(-x)=-xe^x\), which is neither \(f_1(x)\) nor \(-f_1(x)\). Thus, \(f_1\) is neither even nor odd. Also, \(f_1(x)>0\) for \(x>0\), \(f_1(x)<0\) for \(x<0\), and \(f_1(x)\to0\) as \(x\to\infty\). These features match graph c. 2. For \(f_2\), \(f_2(-x)=-f_2(x)\), so \(f_2\) is odd. Its graph has rotational symmetry about the origin and passes through \((-1, 1)\), \((0, 0)\), and \((1, -1)\). These features match graph a. 3. For \(f_3\), \(f_3(-x)=f_3(x)\), so \(f_3\) is even. Its graph is symmetric about the y-axis, has y-intercept \((0, 3)\), and approaches the horizontal asymptote \(y=-1\). These features match graph b.

\(f_1\): neither, graph c; \(f_2\): odd, graph a; \(f_3\): even, graph b

- Compare the effects of even and odd exponents on symmetry. - Determine whether each function can have positive, negative, or all real outputs. - Use the sign of the coefficient to identify whether the even-power graph opens upward or downward.

1. The graph of \(f\) opens downward, has a maximum at \((0, 0)\), and is symmetric about the y-axis because its degree is even. 2. Since \(-0.25x^4\le0\), the range is \((-\infty, 0]\). 3. The function \(g\) has odd degree, so its graph is symmetric about the origin. Its outputs cover all real numbers, so its range is \((-\infty, \infty)\).

1. The graph opens downward, has a maximum at \((0, 0)\), and is symmetric about the y-axis. 2. \((-\infty, 0]\) 3. The graph of \(g\) is symmetric about the origin and has range \((-\infty, \infty)\).

- Apply the even-function definition to a general even exponent. - Ask whether a property shared by every candidate can identify only one of them. - On \(0<x<1\), compare what happens when the exponent becomes larger.

1. For each exponent \(n\in\{2,4,6\}\), \((-x)^n=x^n\) because \(n\) is even. Therefore, each candidate satisfies \(f(-x)=f(x)\) and is an even function. 2. Y-axis symmetry is shared by all three candidates, so it cannot distinguish one candidate from another. A feature common to every option is not enough to identify which option is which. 3. For \(0<x<1\), increasing the positive exponent makes the value smaller. Thus, \(x^2>x^4>x^6\) on that interval. 4. In the graph, \(u\) is highest, \(v\) is in the middle, and \(w\) is lowest between \(0\) and \(1\). Therefore, \(u=x^2\), \(v=x^4\), and \(w=x^6\).

a) All three functions are even. b) Y-axis symmetry cannot distinguish them because all three share that symmetry. c) \(u=x^2\), \(v=x^4\), \(w=x^6\)

- Ask whether one numerical example can establish a statement for every real input. - Apply the defining odd-function equation to an arbitrary \(x\). - Once oddness is proved, use the relationship between outputs at opposite inputs.

1. One pair only verifies the odd-function relation at one input. The definition must hold for every input in the domain. 2. \(f(-x)=0.5(-x)^3=-0.5x^3=-f(x)\) for every real \(x\). Therefore, \(f\) is odd. 3. For an odd function, \(f(-6)=-f(6)=-108\).

a) One pair does not prove an identity that must hold for every input. b) \(f(-x)=-f(x)\) for every real \(x\), so \(f\) is odd. c) \(f(-6)=-108\)

- Write \(h(-x)\) using the even and odd identities. - Recall the value of an odd function at \(0\). - A zero of the sum occurs when the two addends are opposites.

1. Since \(f\) is even and \(g\) is odd, \(h(-x)=f(x)-g(x)\), which is generally not equal to \(h(x)=f(x)+g(x)\). The displayed graph confirms that \(h\) is not symmetric about the y-axis. 2. Every odd function satisfies \(g(0)=0\). The graph shows \(f(0)=0\), so \(h(0)=f(0)+g(0)=0\). 3. A zero of \(h\) satisfies \(f(x)+g(x)=0\), or \(f(x)=-g(x)\). On the graph, look for an x-value where the two function values have equal magnitude and opposite signs. This occurs at \(x=-2.5\).

a) No. In general, \(h(-x)=f(x)-g(x)\ne h(x)\). b) \(h(0)=0\) c) Find where \(f(x)=-g(x)\). The additional zero is \(x=-2.5\).

- Establish the odd-function identity before comparing the two reflections. - Translate each geometric reflection into its function notation. - Part b can be completed directly from \(f(-x)=-f(x)\).

1. \(f(-x)=(-x)^3=-x^3=-f(x)\), so \(f\) is odd. 2. Because \(f\) is odd, \(f(-x)=-f(x)\). Therefore, \(r_y(x)=f(-x)=-f(x)=r_x(x)\) for every real \(x\). 3. Thus, for this odd function, reflecting the graph across the x-axis or across the y-axis produces the same graph. 4. The common reflected graph is \(y=-x^3\). For \(x<0\), the output is positive, and for \(x>0\), the output is negative. Therefore, the graph lies in Quadrants II and IV, apart from the origin.

a) \(f\) is odd. b) \(r_x(x)=r_y(x)\) for every real \(x\). c) The x-axis and y-axis reflections coincide; the reflected graph lies in Quadrants II and IV and passes through the origin.

- Factor out the greatest power of \(x\) from the family expression. - Use the exponent on each factor to determine multiplicity. - Compare the exponents in each expanded function to the definitions of even and odd functions.

a) Factor the family: \(g_n(x)=\frac{1}{10}x^{n-2}(x^2-16)=\frac{1}{10}x^{n-2}(x-4)(x+4)\). For \(n=3\), the zeros \(-4\), \(0\), and \(4\) all have multiplicity \(1\). For \(n=4\), the zero \(0\) has multiplicity \(2\), while \(-4\) and \(4\) each have multiplicity \(1\). b) The function \(g_3(x)=\frac{1}{10}(x^3-16x)\) contains only odd powers, so it is odd. The function \(g_4(x)=\frac{1}{10}(x^4-16x^2)\) contains only even powers, so it is even. c) The function \(g_5(x)=\frac{1}{10}(x^5-16x^3)\) contains only odd powers, so it is odd and its graph is symmetric about the origin.

a) \(g_3\): \(-4\), \(0\), and \(4\), each with multiplicity \(1\). \(g_4\): \(0\) with multiplicity \(2\), and \(-4\) and \(4\) with multiplicity \(1\). b) \(g_3\) is odd; \(g_4\) is even. c) \(g_5\) is odd, so its graph is symmetric about the origin.

- Use an even quartic model with only even powers. - The y-intercept determines the constant term. - Substitute the other two points to form a system.

1. Since \(f\) is an even fourth-degree polynomial, write \(f(x)=ax^4+bx^2+c\). 2. Point \(A(0, 2)\) gives \(c=2\). 3. Point \(B(1, 0)\) gives \(a+b+2=0\), so \(a+b=-2\). 4. Point \(C(2, 0)\) gives \(16a+4b+2=0\), so \(16a+4b=-2\). 5. Substituting \(b=-2-a\) gives \(16a+4(-2-a)=-2\), so \(12a=6\) and \(a=0.5\). Then \(b=-2.5\). 6. Therefore, \(f(x)=0.5x^4-2.5x^2+2\).

\(f(x)=0.5x^4-2.5x^2+2\)

- Use a cubic model containing only odd powers. - Substitute both given points. - Solve the resulting two-equation system.

1. An odd cubic polynomial has the form \(g(x)=ax^3+bx\). 2. Point \(Q(2, 0)\) gives \(8a+2b=0\), or \(4a+b=0\). 3. Point \(P(1, -3)\) gives \(a+b=-3\). 4. Subtracting the second equation from the first gives \(3a=3\), so \(a=1\). Then \(b=-4\). 5. Therefore, \(g(x)=x^3-4x\).

\(g(x)=x^3-4x\)

- Compare y-axis symmetry with origin symmetry. - Determine which graphs pass through the origin and which are undefined at \(x=0\). - Connect the symmetry of \(x^n\) with whether the integer exponent is even or odd.

1. Graph \(r\) is symmetric about the y-axis, so it is an even function and its exponent is even. Graph \(s\) is symmetric about the origin, so it is an odd function and its exponent is odd. Graph \(t\) is symmetric about the y-axis, so it is even. 2. Graph \(r\) passes through the origin and has a positive even exponent, so it represents \(k(x)=x^4\). Graph \(s\) represents \(m(x)=x^3\). Graph \(t\) is undefined at \(x=0\) and has a negative even exponent, so it represents \(p(x)=x^{-2}\).

a) Graph \(r\): even; graph \(s\): odd; graph \(t\): even b) \(r: k(x)=x^4\); \(s: m(x)=x^3\); \(t: p(x)=x^{-2}\)

- Check whether each domain is symmetric about zero. - Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Use zeros and asymptotes to distinguish functions with the same symmetry.

1. The function \(f\) is even because \(f(-x)=f(x)\). Its graph is symmetric about the y-axis, has a maximum at \((0, 3)\), and approaches \(y=0\) as \(x\to\pm\infty\). This matches graph a). 2. The function \(g\) is even because \(g(-x)=g(x)\). It has zeros at \(x=\pm1\) and approaches the horizontal asymptote \(y=1\) as \(|x|\to\infty\). This matches graph b). 3. The function \(h\) is neither even nor odd because its domain is \((0, \infty)\), which is not symmetric about zero. It has a zero at \(x=1\) and a vertical asymptote at \(x=0\). This matches graph c).

a) \(\rightarrow\) 1), even; b) \(\rightarrow\) 2), even; c) \(\rightarrow\) 3), neither

- Count the turning points and inspect the graph's end behavior. - Use the \(180^\circ\) rotational symmetry about the origin. - Write the polynomial in factored form from its zeros. - Substitute the additional point to find the leading factor.

1. The graph has two turning points, so its degree is at least \(3\). A cubic is the least possible degree and is consistent with the graph's end behavior. 2. The graph has \(180^\circ\) rotational symmetry about the origin, so \(g\) is odd. A general cubic with this symmetry is \(g(x)=ax^3+cx\). 3. Using the three zeros, write \(g(x)=a x(x+3)(x-3)=a(x^3-9x)\). Since \(P(1, -2)\) lies on the graph, \(-2=a(1-9)=-8a\), so \(a=\frac{1}{4}\). Therefore, \(g(x)=\frac{1}{4}x^3-\frac{9}{4}x\).

1. Least possible degree: \(3\) 2. \(g(x)=ax^3+cx\) 3. \(g(x)=\frac{1}{4}x^3-\frac{9}{4}x\)

- An even graph mirrors across the y-axis. - An odd graph has 180-degree rotational symmetry about the origin. - Identify the actual center of rotational symmetry for \(h\).

1. The graph of \(f\) is symmetric about the y-axis, so \(f\) is even. 2. The graph of \(g\) is a line through the origin and is symmetric about the origin, so \(g\) is odd. 3. The graph of \(h\) has rotational symmetry about \((0, 2)\), not about the origin. Odd functions must satisfy \(h(-x)=-h(x)\) and, when defined at \(0\), must pass through \((0, 0)\). Therefore, \(h\) is not odd.

a) \(f\) is even. b) \(g\) is odd. c) \(h\) is symmetric about \((0, 2)\), not about the origin, so it is not odd.

- Determine whether the denominator can equal zero. - Find the smallest possible denominator and consider what happens as \(|x|\) becomes large. - Compute \(h(-x)\).

1. Since \(x^2+2\ge2\), the denominator is never zero. Therefore, the domain is \(\mathbb{R}\). 2. The denominator is smallest at \(x=0\), so the maximum output is \(h(0)=3\). The function is always positive, and \(h(x)\to0\) as \(|x|\to\infty\), but it never equals \(0\). Thus, the range is \((0, 3]\). 3. Replace \(x\) with \(-x\): \(h(-x)=\frac{6}{(-x)^2+2}=\frac{6}{x^2+2}=h(x)\). Therefore, \(h\) is even and its graph is symmetric about the y-axis.

a) Domain: \(\mathbb{R}\); range: \((0, 3]\) b) \(h(-x)=h(x)\), so \(h\) is even.

- Distinguish a graph being symmetric with itself from two different graphs being mirror images. - Write the algebraic condition for each relationship separately. - A counterexample only needs two reflected functions for which neither function has y-axis symmetry by itself.

1. The claim is false. 2. A single graph is symmetric about the y-axis when its domain is symmetric about \(0\) and \(f(-x)=f(x)\) for every \(x\) in the domain. 3. Two graphs are reflections of each other across the y-axis when their domains are reflected across \(0\) and \(g(x)=f(-x)\). 4. Let \(f(x)=x+1\) and \(g(x)=-x+1\). Since \(f(-x)=-x+1=g(x)\), the graphs are reflections of each other. However, \(f(-x)\ne f(x)\) and \(g(-x)\ne g(x)\) in general, so neither graph is individually symmetric about the y-axis.

The claim is false. Individual y-axis symmetry requires \(f(-x)=f(x)\). Two graphs are reflections across the y-axis when \(g(x)=f(-x)\), with correspondingly reflected domains. For example, \(f(x)=x+1\) and \(g(x)=-x+1\) are reflections of each other, but neither function is even.

- Define the shifted function and test both symmetry identities. - Compare \(g(-x)\) with \(-g(x)\). - For even symmetry, determine what \(-f(x)=f(x)\) would imply.

1. Let \(f\) be a nonzero odd polynomial, so \(f(-x)=-f(x)\), and define \(g(x)=f(x)+c\) with \(c\ne0\). 2. For odd symmetry, \(g(-x)=-f(x)+c\), while \(-g(x)=-f(x)-c\). These are equal only if \(c=0\), so \(g\) is not odd. 3. For even symmetry, \(g(-x)=g(x)\) would require \(-f(x)+c=f(x)+c\), or \(f(x)=0\) for every \(x\). 4. That would make \(f\) the zero polynomial, contrary to the hypothesis. Therefore, \(g\) is not even. 5. The statement is true.

The statement is true. A nonzero vertical shift destroys odd symmetry, and the shifted function could be even only if the original odd polynomial were the zero polynomial.

- Begin with \(f(-x) = f(x)\). - Test the definitions of even and odd functions for parts a) and b). - For part c), remember that a nonconstant even polynomial has even degree. - Consider what the horizontal shift does to the next-highest power in the leading term.

1. Because \(f\) is even, \(f(-x) = f(x)\). 2. For \(p\), \(p(-x) = (-x)f(-x) = -xf(x) = -p(x)\). Therefore, \(p\) is odd. 3. For \(q\), \(q(-x) = f(-x) + (-x)^2 = f(x) + x^2 = q(x)\). Therefore, \(q\) is even. 4. Since \(f\) is a nonconstant even polynomial, its degree is an even number \(2n\). The translated polynomial \(r(x) = f(x + 1)\) still has even degree, so it cannot be odd. 5. If the leading term of \(f\) is \(ax^{2n}\), then expanding \(a(x + 1)^{2n}\) produces the nonzero odd-power term \(2na x^{2n-1}\). Lower-degree terms cannot cancel it. Therefore, \(r\) is not even. Thus, \(r\) is neither even nor odd.

a) Odd; the graph is symmetric about the origin. b) Even; the graph is symmetric about the y-axis. c) Neither even nor odd.

- Write the horizontal recentering as a single shift parameter before expanding. - An odd polynomial cannot contain a nonzero even-power term or a nonzero constant term. - Once the recentered function is odd, interpret the horizontal and vertical recentering geometrically.

1. Let \(t=a-5\). Then \(h(u)=(u+t)^3+4(u+t)-2-b\). 2. Expanding gives \(h(u)=u^3+3tu^2+(3t^2+4)u+(t^3+4t-2-b)\). 3. For \(h\) to be odd, the even-power term and the constant term must be zero. The coefficient of \(u^2\) gives \(3t=0\), so \(t=0\) and therefore \(a=5\). 4. With \(t=0\), the constant term is \(-2-b\). Setting it equal to zero gives \(b=-2\). 5. Thus \(h(u)=u^3+4u\), which is odd. Recentring the graph at \((a,b)=(5,-2)\) produces an odd graph about the origin, so the original graph has center of symmetry \((5, -2)\).

\(a=5\), \(b=-2\). Then \(h(u)=u^3+4u\) is odd, so the graph of \(f\) has center of symmetry \((5, -2)\).