Free Math Worksheets

- Start with familiar power functions. - Consider how the sign of the leading coefficient affects the graph. - For part c, look for a function that is increasing across its entire domain.

1. A quadratic with a positive leading coefficient opens upward and has one local minimum. One example is \(h(x)=x^2\). 2. The fourth-degree polynomial \(h(x)=-x^4\) has exactly one local maximum, at the origin. 3. A strictly increasing function has no local extrema. One example is \(h(x)=x^3\).

Possible answers: a) \(h(x)=x^2\) b) \(h(x)=-x^4\) c) \(h(x)=x^3\)

- Identify the symmetry from the powers in the polynomial. - Reflect the positive x-coordinates across the y-axis. - Use the positive leading term and even degree to determine what happens on both ends. - Local extrema alternate in type along a smooth polynomial graph.

1. The polynomial contains only even powers, so it is even and its graph is symmetric about the y-axis. 2. Therefore, the extrema at \(x=1\) and \(x=2\) have matching extrema of the same type at \(x=-1\) and \(x=-2\). 3. The leading term is \(\frac{1}{6}x^6\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). Thus, the outer extrema at \(x=\pm2\) are local minima. 4. The extrema alternate in type, so \(x=\pm1\) are local maxima and \(x=0\) is a local minimum.

a) \(x=-1\) and \(x=-2\) b) Local minima: \(x=-2, 0, 2\) Local maxima: \(x=-1, 1\)

- Identify the symmetry from the odd powers. - Reflect each extremum through the origin. - Determine what happens to a peak under a 180-degree rotation.

1. The polynomial contains only odd powers, so \(g\) is odd and its graph is symmetric about the origin. 2. The extrema at \(x=1\) and \(x=2\) are reflected to extrema at \(x=-1\) and \(x=-2\). 3. A 180-degree rotation about the origin changes a local maximum into a local minimum and a local minimum into a local maximum. 4. Therefore, \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

a) \(x=-1\) and \(x=-2\) b) \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

- Rewrite the quadratic in vertex form. - Express both vertex coordinates in terms of \(k\). - Eliminate \(k\) and include the restriction implied by \(k > 0\).

1. Rewrite the function in vertex form: \(f_k(x) = -\frac{2}{k}(x - 2k)^2 + 8k + 1\). 2. Thus, the vertex has coordinates \((2k, 8k + 1)\). 3. From \(x = 2k\), obtain \(k = \frac{x}{2}\). Substitute into the \(y\)-coordinate: \(y = 8\left(\frac{x}{2}\right) + 1 = 4x + 1\). 4. Because \(k > 0\), the vertex satisfies \(x > 0\). Therefore, the locus is \(y = 4x + 1\), for \(x > 0\).

\(y = 4x + 1\), for \(x > 0\)

- Local maxima and minima alternate. - Follow the only possible shape through the extrema from left to right. - Use the fact that the function is a polynomial and has no other extrema. - Count x-axis crossings from the end behavior and the positive extremum values.

1. Consecutive local extrema alternate between maxima and minima. Since the function value decreases from \(A\) to \(B\) and then increases from \(B\) to \(C\), \(A\) and \(C\) are local maxima, while \(B\) is a local minimum. 2. To the left of the leftmost local maximum, the polynomial must increase toward \(A\), so \(f(x)\to-\infty\) as \(x\to-\infty\). To the right of the rightmost local maximum, it must decrease away from \(C\), so \(f(x)\to-\infty\) as \(x\to\infty\). 3. All three extremum values are positive. The graph rises from \(-\infty\) to \(A\), so it crosses the x-axis exactly once before \(x=-2\). Between the extrema, the graph remains above the x-axis. After \(C\), it falls to \(-\infty\), so it crosses exactly once more. Therefore, \(f\) has exactly two real zeros.

a) \(A\) and \(C\) are local maxima; \(B\) is a local minimum. b) \(f(x)\to-\infty\) as \(x\to\pm\infty\). c) Exactly two real zeros.

- Substitute both given points into \(f(x) = ax^2 + bx + c\). - A downward-opening parabola has a negative leading coefficient. - Use the vertex formula \(x_v = -\frac{b}{2a}\) and require \(x_v > 0\).

1. Start with \(f(x) = ax^2 + bx + c\). Since the graph passes through \((0, 1.5)\), \(c = 1.5\). 2. Since the graph also passes through \((30, 0)\), \(900a + 30b + 1.5 = 0\). Solving for \(b\) gives \(b = -30a - 0.05\). Therefore, \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\). 3. The parabola opens downward when \(a < 0\). Its vertex has \(x\)-coordinate \(x_v = -\frac{b}{2a} = \frac{30a + 0.05}{2a}\). 4. Requiring \(x_v > 0\), while \(a < 0\), gives \(30a + 0.05 < 0\). Thus, \(a < -\frac{1}{600}\). 5. The maximum height occurs at the vertex, whose \(x\)-coordinate is \(x_v = \frac{30a + 0.05}{2a} = 15 + \frac{0.025}{a}\).

a) \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\) b) \(a < -\frac{1}{600}\) c) \(x = 15 + \frac{0.025}{a}\) feet from the nozzle

- Identify each peak and valley. - Compare the y-values of all extrema. - Include the endpoint values when finding absolute extrema on a closed interval. - A local extremum need not be the highest or lowest point on the entire interval.

1. From the graph, the local maxima are \((-2.5, 3)\) and \((1, 4)\), and the local minima are \((-1, 1)\) and \((3, -3)\). 2. The endpoint values are \(f(-4)=-2\) and \(f(4)=-1\). The largest value on the interval is \(4\), so \((1, 4)\) is the absolute maximum. The smallest value is \(-3\), so \((3, -3)\) is the absolute minimum. The other two extrema are local only.

a) Local maxima: \((-2.5, 3)\), \((1, 4)\); local minima: \((-1, 1)\), \((3, -3)\) b) Absolute maximum: \((1, 4)\); absolute minimum: \((3, -3)\). The points \((-2.5, 3)\) and \((-1, 1)\) are local only.

- Interpret \(f(x)=c\) as intersections with a horizontal line \(y=c\). - Identify the y-values of the local maximum and the two global minima. - Check what happens when the horizontal line passes exactly through an extremum. - Consider the graph's end behavior beyond the visible turning points.

1. Solutions to \(f(x)=c\) are the x-coordinates where the horizontal line \(y=c\) intersects the graph. 2. For \(c=0\), the graph crosses the x-axis four times, so there are four solutions. 3. The graph has two global minima at height \(-4\) and a local maximum at height \(6\). A horizontal line intersects all four branches when \(-4<c<6\). 4. At \(c=-4\), the line is tangent at the two global minima, giving exactly two solutions. For \(c>6\), the line intersects only the two outer branches, also giving exactly two solutions. 5. At \(c=6\), the line passes through the local maximum and intersects both outer branches, giving exactly three solutions.

a) \(4\) solutions b) \(-4<c<6\) c) \(c=-4\) or \(c>6\) d) \(c=6\)

- Complete the square to identify the vertex. - Compare the vertex's \(y\)-coordinate with zero. - Substitute the coordinates of the proposed common point into the equation.

1. Complete the square: \(g_k(x) = \left(x - \frac{k}{2}\right)^2 - \frac{k^2}{4} + 2k - 3\). Therefore, the vertex is \(\left(\frac{k}{2}, -\frac{k^2}{4} + 2k - 3\right)\). 2. Because each parabola opens upward, the number of real zeros depends on the vertex's \(y\)-coordinate. Factor its negative: \(-\frac{k^2}{4} + 2k - 3 = -\frac{1}{4}(k - 2)(k - 6)\). 3. The vertex is below the \(x\)-axis when \(k < 2\) or \(k > 6\), so the graph has two real zeros. It is on the \(x\)-axis when \(k = 2\) or \(k = 6\), so the graph has one real zero. It is above the \(x\)-axis when \(2 < k < 6\), so the graph has no real zeros. 4. Substitute \(x = 2\): \(g_k(2) = 2^2 - 2k + 2k - 3 = 1\). This value is independent of \(k\), so \((2, 1)\) lies on every graph.

a) \(g_k(x) = \left(x - \frac{k}{2}\right)^2 - \frac{k^2}{4} + 2k - 3\); vertex \(\left(\frac{k}{2}, -\frac{k^2}{4} + 2k - 3\right)\) b) Two real zeros for \(k < 2\) or \(k > 6\); one real zero for \(k = 2\) or \(k = 6\); no real zeros for \(2 < k < 6\) c) \(g_k(2) = 1\) for every \(k\).

- Complete the square to find the vertex. - Eliminate the parameter from the vertex coordinates. - Compare the vertex's \(y\)-coordinate with zero to count the real zeros. - Collect the terms containing \(a\) to find where the function value is independent of the parameter.

1. Complete the square: \(f_a(x) = (x - a)^2 - a^2 + 3a\). Therefore, the vertex is \((a, -a^2 + 3a)\). 2. At a vertex, \(x = a\). Replacing \(a\) with \(x\) in the \(y\)-coordinate gives the locus \(y = -x^2 + 3x\). 3. Each parabola opens upward, so the number of real zeros depends on the vertex's \(y\)-coordinate \(-a^2 + 3a = a(3 - a)\). It is negative for \(a < 0\) or \(a > 3\), zero for \(a = 0\) or \(a = 3\), and positive for \(0 < a < 3\). 4. Thus, there are two real zeros for \(a < 0\) or \(a > 3\), one real zero for \(a = 0\) or \(a = 3\), and no real zeros for \(0 < a < 3\). 5. Rewrite \(f_a(x) = x^2 + a(3 - 2x)\). For the function value to be independent of \(a\), require \(3 - 2x = 0\), so \(x = \frac{3}{2}\). Then \(y = x^2 = \frac{9}{4}\). The common point is \(\left(\frac{3}{2}, \frac{9}{4}\right)\).

a) \(f_a(x) = (x - a)^2 - a^2 + 3a\); vertex: \((a, -a^2 + 3a)\) b) \(y = -x^2 + 3x\) c) Two real zeros for \(a < 0\) or \(a > 3\); one real zero for \(a = 0\) or \(a = 3\); no real zeros for \(0 < a < 3\) d) \(\left(\frac{3}{2}, \frac{9}{4}\right)\)

- Complete the square to identify the vertex. - Eliminate \(t\) from the vertex coordinates. - When does the vertex of a downward-opening parabola lie on the \(x\)-axis? - Collect the terms containing \(t\) to find a parameter-independent point.

1. Complete the square: \(h_t(x) = -\left(x - \frac{t}{2}\right)^2 + \frac{t^2}{4} - 2t + 4 = -\left(x - \frac{t}{2}\right)^2 + \frac{(t - 4)^2}{4}\). 2. Therefore, the vertex is \(\left(\frac{t}{2}, \frac{(t - 4)^2}{4}\right)\). 3. From \(x = \frac{t}{2}\), obtain \(t = 2x\). Substitution into the vertex's \(y\)-coordinate gives \(y = \frac{(2x - 4)^2}{4} = (x - 2)^2\). 4. A downward-opening parabola is tangent to the \(x\)-axis when its vertex lies on the axis. Thus, \(\frac{(t - 4)^2}{4} = 0\), which gives \(t = 4\). 5. Rewrite \(h_t(x) = -x^2 + 4 + t(x - 2)\). At \(x = 2\), the parameter term is zero and \(h_t(2) = 0\). Therefore, every graph passes through \((2, 0)\).

a) \(h_t(x) = -\left(x - \frac{t}{2}\right)^2 + \frac{(t - 4)^2}{4}\); vertex: \(\left(\frac{t}{2}, \frac{(t - 4)^2}{4}\right)\) b) \(y = (x - 2)^2\) c) \(t = 4\) d) \((2, 0)\)

- Use the vertex formula for a quadratic. - Express the vertex coordinates in terms of \(a\), then eliminate the parameter. - Determine when the parabola opens downward. - Account for the sign of \(a\) when solving the inequality for the vertex's height.

1. The vertex formula gives \(x_v = -\frac{-4}{2a} = \frac{2}{a}\). 2. Evaluate the function at the vertex: \(y_v = f_a\left(\frac{2}{a}\right) = a - \frac{4}{a}\). 3. From \(x = \frac{2}{a}\), obtain \(a = \frac{2}{x}\). Substituting into the \(y\)-coordinate gives \(y = \frac{2}{x} - 2x\). Since \(a \ne 0\), \(x \ne 0\). 4. The vertex is a maximum when the parabola opens downward, so \(a < 0\). 5. It lies below the \(x\)-axis when \(a - \frac{4}{a} < 0\). Because \(a < 0\), multiplying by \(a\) reverses the inequality: \(a^2 - 4 > 0\). Combined with \(a < 0\), this gives \(a < -2\).

a) \(y = \frac{2}{x} - 2x\), for \(x \ne 0\) b) \(a < -2\)

- Compare the two ends of an even-degree polynomial graph. - Use continuity to reason about attaining a highest or lowest value. - The derivative of a degree-\(4\) polynomial has degree \(3\). - Local extrema can occur only at critical values.

1. For an even-degree polynomial, both ends of the graph approach \(\infty\) when the leading coefficient is positive, or both approach \(-\infty\) when it is negative. By continuity, the positive-leading case attains an absolute minimum, and the negative-leading case attains an absolute maximum. 2. The derivative of a quartic polynomial is a cubic polynomial. A cubic can have at most three real zeros. Since every local extremum of the quartic must occur at a zero of its derivative, the quartic can have at most three local extrema.

a) Every even-degree polynomial has an absolute minimum when its leading coefficient is positive or an absolute maximum when its leading coefficient is negative. b) Its derivative is cubic and has at most three real zeros, so the quartic has at most three local extrema.

- Interpret \(g(x)=k\) as intersections with a horizontal line \(y=k\). - Track what changes when the line passes through the minimum, the x-axis, and the maximum. - Remember that the right branch approaches \(y=0\) but does not reach it again.

1. Zeros are x-intercepts. The graph crosses the x-axis at \(x=-3\) and \(x=0\), so \(g\) has two zeros. 2. A horizontal line \(y=k\) intersects the graph twice when \(-2<k\le0\). It also intersects twice when \(k=2\): once on the left branch and once at the local maximum. 3. When \(0<k<2\), the horizontal line intersects the left branch once and the right-hand hump twice, for three solutions. 4. When \(k<-2\), the horizontal line lies below the global minimum and does not intersect the graph. At \(k=-2\), there is one solution.

a) \(2\) b) \(k\in(-2, 0]\cup\{2\}\) c) \(k\in(0, 2)\) d) \(k<-2\)