Free Math Worksheets

- What does the derivative represent at a point? - Differentiate each term of the function. - Substitute \(x=5\) into the derivative.

1. Differentiate the function: \(y'=0.4x+5\). 2. Evaluate the derivative at \(x=5\): \(y'(5)=0.4\cdot 5+5\). 3. Simplify: \(y'(5)=2+5=7\).

\(7\)

- What derivative value corresponds to a horizontal tangent? - Set the derivative equal to the required slope. - After finding an x-coordinate, substitute it into the original function to find the y-coordinate.

1. Differentiate using the power, constant, and sum rules: \(f'(x)=-2x+4\). 2. A horizontal tangent has slope \(0\), so solve \(-2x_0+4=0\). This gives \(x_0=2\). 3. Evaluate the function: \(f(2)=-(2)^2+4\cdot 2-2=2\). Therefore, \(P=(2,2)\). 4. For slope \(6\), solve \(-2q+4=6\). This gives \(q=-1\). 5. Evaluate the function: \(f(-1)=-(-1)^2+4(-1)-2=-7\). Therefore, \(Q=(-1,-7)\).

a) \(f'(x)=-2x+4\) b) \(P=(2,2)\) c) \(Q=(-1,-7)\)

- Differentiate each term separately using the power, constant-multiple, and sum rules. - What is the derivative of a term that does not contain the variable? - Treat \(k\) and \(n\) as parameters and \(\pi\) as a constant.

1. Differentiate each term in part a): \(f'(x)=\frac{1}{4}(4x^3)-2(3x^2)+5=x^3-6x^2+5\). 2. Differentiate each term in part b). The constant term has derivative \(0\): \(f'(x)=\frac{2}{3}(6x^5)+2\sqrt{5}x=4x^5+2\sqrt{5}x\). 3. Treat \(k\) and \(n\) as constants in part c): \(f'(x)=knx^{n-1}+2\).

a) \(f'(x)=x^3-6x^2+5\) b) \(f'(x)=4x^5+2\sqrt{5}x\) c) \(f'(x)=knx^{n-1}+2\)

- Rewrite each expression as a sum of simple power functions before differentiating. - Expand the squared binomial in part b). - Identify the variable of differentiation and treat every parameter as a constant.

1. Rewrite part a) as \(f(x)=\frac{1}{3}x^3-2x\). Then \(f'(x)=x^2-2\). 2. Expand part b): \(f(x)=x^2+6x+9\). Therefore, \(f'(x)=2x+6\). 3. In part c), treat \(a\) as a constant. Then \(f'(x)=2a^2x+a\).

a) \(f'(x)=x^2-2\) b) \(f'(x)=2x+6\) c) \(f'(x)=2a^2x+a\)

- Use the constant-multiple rule when a coefficient multiplies a power of \(x\). - In the power rule, how does the exponent change? - What is the derivative of a constant? - Differentiate a sum term by term.

1. Apply the power rule to the first term: \(\frac{d}{dx}\left(\frac{1}{2}x^4\right)=2x^3\). 2. Differentiate the second term: \(\frac{d}{dx}(-4x^3)=-12x^2\). 3. The derivative of \(12x\) is \(12\), and the derivative of \(-5\) is \(0\). 4. Combine the results: \(f'(x)=2x^3-12x^2+12\).

\(f'(x)=2x^3-12x^2+12\)

- Differentiate only with respect to the variable shown in the function notation. - A constant term has derivative \(0\). - For the power rule, multiply by the exponent and then decrease the exponent by \(1\). - Treat parameters as constants.

1. For part a), apply the power rule term by term: \(f'(x)=x^4-6x\). 2. For part b), differentiate with respect to \(t\): \(f'(t)=t^3-2t^2+7\). 3. For part c), treat \(m\), \(n\), and \(p\) as constants: \(f'(x)=2mx+n\). 4. For part d), differentiate with respect to \(k\) and treat \(x^3\) as a constant coefficient: \(f'(k)=2x^3k-5\).

a) \(f'(x)=x^4-6x\) b) \(f'(t)=t^3-2t^2+7\) c) \(f'(x)=2mx+n\) d) \(f'(k)=2x^3k-5\)

- For each term, decide whether it contains the variable of differentiation. - A term that does not contain the variable has derivative \(0\). - Simplify numerical coefficients after applying the power rule. - In part d), \(x\) is the variable and \(k\) is a parameter.

1. For part a), differentiate term by term: \(f'(z)=\frac{1}{2}z^2-2z+\frac{1}{2}\). 2. In part b), treat \(a\) as a constant: \(f'(x)=28x^6-6ax^2\). 3. In part c), \(v\) and \(s_0\) are constants, so \(s'(t)=v\). 4. In part d), treat \(k\) as a constant. Since \(-5k\) does not contain \(x\), its derivative is \(0\), and \(f'(x)=3k^2x^2\).

a) \(f'(z)=\frac{1}{2}z^2-2z+\frac{1}{2}\) b) \(f'(x)=28x^6-6ax^2\) c) \(s'(t)=v\) d) \(f'(x)=3k^2x^2\)

- Differentiate a sum one term at a time. - Apply the power rule to both the exponent and coefficient. - A constant has derivative \(0\). - Treat a decimal coefficient the same way as any other constant coefficient.

1. Differentiate each power term: \(\frac{d}{dx}\left(\frac{5}{6}x^6\right)=5x^5\), \(\frac{d}{dx}(-2x^3)=-6x^2\), and \(\frac{d}{dx}(0.4x^2)=0.8x\). 2. The derivative of \(-9\) is \(0\). 3. Combine the terms: \(f'(x)=5x^5-6x^2+0.8x\).

\(f'(x)=5x^5-6x^2+0.8x\)

- What role does \(a\) play when differentiating with respect to \(t\)? - The power rule still applies when a coefficient contains a parameter. - What is the derivative of the linear term \(5t\)? - Since \(a\) is part of a constant coefficient, it remains in the denominator.

1. Differentiate with respect to \(t\), treating \(a\) as a constant. 2. The first term gives \(at^2\), the second gives \(-\frac{2}{a}t\), and the third gives \(5\). 3. Therefore, \(g_a'(t)=at^2-\frac{2}{a}t+5\).

\(g_a'(t)=at^2-\frac{2}{a}t+5\)

- What is the derivative of a term of the form \(cx\)? - How does the exponent change under the power rule? - Check every term of the original function separately.

1. The student left the variable in the derivative of the linear term. Since \(\frac{d}{dx}(x)=1\), the derivative of \(7x\) is \(7\), not \(7x\). 2. Differentiate all terms: \(\frac{d}{dx}\left(\frac{1}{3}x^3\right)=x^2\), \(\frac{d}{dx}(-4x^2)=-8x\), \(\frac{d}{dx}(7x)=7\), and \(\frac{d}{dx}(-5)=0\). 3. Therefore, \(f'(x)=x^2-8x+7\).

The derivative of \(7x\) is \(7\), not \(7x\). The correct derivative is \(f'(x)=x^2-8x+7\).

- What is the derivative of a term that does not contain \(x\)? - Rewrite a power of \(x\) in the denominator using a negative exponent. - When applying the power rule to a negative exponent, check the signs in the coefficient multiplication.

1. In (1), the constant rule was applied incorrectly. The derivative of \(12\) is \(0\), so \(g'(x)=2x^3-4x\). 2. In (2), rewrite \(-\frac{3}{x^2}\) as \(-3x^{-2}\). Its derivative is \((-3)(-2)x^{-3}=6x^{-3}\), so \(h'(x)=10x^4+6x^{-3}=10x^4+\frac{6}{x^3}\).

(1) The constant \(12\) should have derivative \(0\). The correct derivative is \(g'(x)=2x^3-4x\). (2) The sign is wrong because \((-3)(-2)=6\). The correct derivative is \(h'(x)=10x^4+6x^{-3}\), or \(h'(x)=10x^4+\frac{6}{x^3}\).

- Substitute the given x-value into the function to find the point of tangency. - The derivative gives the slope of the tangent line. - Use point-slope form after finding the point and slope. - A tangent line matches the curve's instantaneous slope at the point of tangency.

1. Find the point of tangency: \(f(1)=1^3-3(1)^2+2=0\), so the point is \((1,0)\). 2. Differentiate: \(f'(x)=3x^2-6x\). 3. Find the tangent slope: \(f'(1)=3(1)^2-6(1)=-3\). 4. Use point-slope form: \(y-0=-3(x-1)\). 5. Simplify: \(y=-3x+3\).

\(y=-3x+3\)

- First find the point through which the normal line passes. - Tangent and normal slopes are negative reciprocals. - Use the derivative to find the tangent slope. - Substitute the slope and point into a line equation.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4+4=4\), so the point is \((4,4)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(f'(4)=2\). 3. The normal slope is the negative reciprocal: \(m_n=-\frac{1}{2}\). 4. Use point-slope form: \(y-4=-\frac{1}{2}(x-4)\), so \(y=-0.5x+6\).

\(n: y=-0.5x+6\)

- A line is determined by its slope and one point. - Use the derivative to find the curve's slope at \(x=2\). - Substitute \(x=2\) into the original function to find the point of tangency. - Use point-slope form to write the line equation.

1. Find the point of tangency: \(f(2)=0.5\cdot 2^4-3\cdot 2^2+4=0\), so the point is \((2,0)\). 2. Differentiate: \(f'(x)=2x^3-6x\). 3. Find the tangent slope: \(f'(2)=2\cdot 2^3-6\cdot 2=4\). 4. Use point-slope form: \(y-0=4(x-2)\). 5. Simplify: \(y=4x-8\).

\(y=4x-8\)

- The derivative at a point gives the tangent line's slope. - How are a line's slope \(m\) and its angle \(\alpha\) related? - Use degree mode when evaluating the inverse tangent.

1. Differentiate: \(f'(x)=-1.5x^2+4x\). 2. Find the tangent slope at \(x=1\): \(m=f'(1)=-1.5(1)^2+4(1)=2.5\). 3. The angle \(\alpha\) satisfies \(\tan(\alpha)=m\), so \(\alpha=\arctan(2.5)\). 4. In degree mode, \(\alpha\approx 68.2^\circ\).

\(68.2^\circ\)

- The derivative at a point gives the tangent line's slope. - Use point-slope form to write the tangent line. - Relate a line's slope to its angle with the positive x-axis.

1. Find the point of tangency: \(f(3)=\frac{1}{3}\cdot 3^3-3^2-2\cdot 3=-6\), so the point is \((3,-6)\). 2. Differentiate: \(f'(x)=x^2-2x-2\). 3. Find the slope: \(f'(3)=3^2-2\cdot 3-2=1\). 4. Use point-slope form: \(y+6=x-3\), so \(y=x-9\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

The tangent line is \(y=x-9\), and its angle with the positive x-axis is \(45^\circ\).

- A line is determined by a point and a slope. - Use the derivative to find the tangent slope. - First calculate the complete point of tangency. - Use point-slope form to write the line.

1. Find the point of tangency: \(f(1)=1^3-4(1)^2+5=2\), so \(P=(1,2)\). 2. Differentiate: \(f'(x)=3x^2-8x\). 3. Find the tangent slope: \(f'(1)=3(1)^2-8(1)=-5\). 4. Use point-slope form: \(y-2=-5(x-1)\). 5. Simplify: \(y=-5x+7\).

\(y=-5x+7\)

- Evaluate the function to find the point on the graph. - Tangent and normal slopes are negative reciprocals. - Use point-slope form after finding the normal slope.

1. Find the point: \(f(3)=3^2-4\cdot3+1=-2\), so the point is \((3,-2)\). 2. Differentiate: \(f'(x)=2x-4\). The tangent slope is \(f'(3)=2\). 3. The normal slope is \(-\frac{1}{2}\). 4. Use point-slope form: \(y+2=-\frac{1}{2}(x-3)\), so \(y=-0.5x-0.5\).

\(n: y=-0.5x-0.5\)

- Use the derivative to find the tangent slope at the given point. - The normal line is perpendicular to the tangent line. - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Find the point: \(f(2)=\frac{1}{4}\cdot2^3+2=4\), so \(P=(2,4)\). 2. Differentiate: \(f'(x)=\frac{3}{4}x^2+1\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y-4=-\frac{1}{4}(x-2)\), so \(y=-0.25x+4.5\).

\(n: y=-0.25x+4.5\)

- What must be true about the function values for two graphs to meet at a point? - What must also be true about their derivatives for the graphs to be tangent there? - Use point-slope form for the common tangent line.

1. The function values agree: \(f(1)=1+1+1=3\) and \(g(1)=-1+5-1=3\). Thus, both graphs pass through \((1,3)\). 2. Differentiate: \(f'(x)=2x+1\) and \(g'(x)=-2x+5\). 3. The slopes also agree at \(x=1\): \(f'(1)=3\) and \(g'(1)=3\). Therefore, the graphs are tangent at \((1,3)\). 4. The common tangent line has slope \(3\) and passes through \((1,3)\): \(y-3=3(x-1)\), so \(y=3x\).

1. Both graphs pass through \((1,3)\), and \(f'(1)=g'(1)=3\), so they are tangent there. 2. \(y=3x\)

- Evaluate the function to find the point on the graph. - Use the derivative for the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Write the line using its slope and the point.

1. Find the point: \(f(2)=\frac{1}{2}\cdot2^3-2\cdot2=0\), so the point is \((2,0)\). 2. Differentiate: \(f'(x)=\frac{3}{2}x^2-2\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y=-\frac{1}{4}(x-2)\), so \(y=-0.25x+0.5\).

\(n: y=-0.25x+0.5\)

- Identify the variable and treat all parameters as constants. - A constant term has derivative \(0\). - Apply the constant-multiple rule to each power term. - Differentiate a sum term by term.

1. For part a), differentiate term by term: \(f'(x)=2x^4-6x^2\). 2. For part b), \(g'(t)=\frac{1}{2}t^2-\frac{3}{2}t+1\). 3. For part c), treat \(k\) as a constant: \(h'(x)=2k^2x-2k\).

a) \(f'(x)=2x^4-6x^2\) b) \(g'(t)=\frac{1}{2}t^2-\frac{3}{2}t+1\) c) \(h'(x)=2k^2x-2k\)

- Can you rewrite the expression as a sum of simple powers before differentiating? - A term with no variable has derivative \(0\). - Fractions and symbols such as \(\pi\) can be constant coefficients or constant terms.

1. Expand part a): \(f(x)=x^3+2x^2-5x-10\). Then \(f'(x)=3x^2+4x-5\). 2. Differentiate part b) term by term. Since \(\pi\) is constant, \(g'(x)=x^3-2x^2\). 3. In part c), differentiate with respect to \(z\), treating \(c\) and \(\frac{1}{c}\) as constants: \(h'(z)=3cz^2-\frac{1}{c}\).

a) \(f'(x)=3x^2+4x-5\) b) \(g'(x)=x^3-2x^2\) c) \(h'(z)=3cz^2-\frac{1}{c}\)

- Parallel lines have equal slopes. - Set the derivative equal to the slope of the given line. - Find the point of tangency after finding its x-coordinate. - Use point-slope form to write the line.

1. Differentiate: \(f'(x)=0.5x+1\). 2. A parallel tangent must have slope \(2\), so solve \(0.5x+1=2\). This gives \(x=2\). 3. Find the point of tangency: \(f(2)=0.25(2)^2+2-3=0\), so the point is \((2,0)\). 4. Write the tangent line: \(y-0=2(x-2)\), so \(y=2x-4\).

\(y=2x-4\)

- Substitute a point into the function to require that it lies on the graph. - Differentiate while treating the parameter as a constant. - A specified tangent slope gives an equation involving the derivative. - Notice whether the derivative at a given input depends on the parameter.

1. For the graph to pass through \((2,5)\), require \(5=4a-8+1\). Thus, \(4a=12\), so \(a=3\). 2. Differentiate: \(f_a'(x)=2ax-4\). At \(x=0\), \(f_a'(0)=-4\), independent of \(a\). 3. For slope \(10\) at \(x=1\), solve \(f_a'(1)=10\): \(2a-4=10\). Thus, \(a=7\).

a) \(a=3\) b) \(-4\) c) \(a=7\)

- Apply the power rule to each term separately. - What happens to a constant term that does not contain \(x\)? - Treat decimal and fractional coefficients as constant factors.

1. Differentiate each term using the power, constant-multiple, and sum rules. 2. The constant term has derivative \(0\), so \(f'(x)=x^4-3x^3+7\).

\(f'(x)=x^4-3x^3+7\)

- Identify the variable of differentiation and treat the other letters as constants. - A parameter multiplying a power remains as a constant coefficient. - A term without the variable has derivative \(0\). - Expanding the squared binomial may simplify part b).

1. In part a), differentiate with respect to \(x\) and treat \(a\) as a constant: \(h'(x)=\frac{3}{a}x^2+2ax-a^3\). 2. For part b), expand first: \(p(t)=9t^2-6kt+k^2\). 3. Differentiate with respect to \(t\), treating \(k\) as a constant: \(p'(t)=18t-6k\).

a) \(h'(x)=\frac{3}{a}x^2+2ax-a^3\) b) \(p'(t)=18t-6k\)

- Decide which symbol is the variable and which is treated as a constant. - Differentiate the sum term by term. - What is the derivative of a term that contains no \(x\)?

1. Treat \(k\) as a constant and differentiate each term. 2. By the power rule, \(\frac{d}{dx}\left(\frac{1}{2k}x^4\right)=\frac{2}{k}x^3\). 3. Also, \(\frac{d}{dx}(-k^2x^2)=-2k^2x\), and \(\frac{d}{dx}(7k)=0\). 4. Therefore, \(f_k'(x)=\frac{2}{k}x^3-2k^2x\).

\(f_k'(x)=\frac{2}{k}x^3-2k^2x\)

- The derivative gives the tangent slope. - Use the tangent function to relate slope and angle. - Find the point of tangency from the original function. - Use point-slope form to write the tangent line.

1. Differentiate: \(f'(x)=-x\). Therefore, the slope at \(x=1\) is \(f'(1)=-1\). 2. Since \(\tan(\alpha)=-1\) and \(\alpha\in[0^\circ,180^\circ)\), \(\alpha=135^\circ\). 3. The point of tangency is \((1,f(1))=(1,2.5)\). Using slope \(-1\), \(y-2.5=-(x-1)\), so \(y=-x+3.5\).

1. \(-1\) 2. \(135^\circ\) 3. \(y=-x+3.5\)

- Expanding a product can turn it into a sum of powers. - A constant term has derivative \(0\). - The value of the derivative at a given input is the graph's slope there.

1. For part a), differentiate term by term: \(f'(x)=3x^5-12x^2+2\). 2. For part b), the constant \(\pi\) has derivative \(0\), so \(f'(x)=2x^2-x\). 3. For part c), expand: \(f(x)=2x^3-x^2\). Then \(f'(x)=6x^2-2x\). 4. Evaluate at \(x=2\): \(f'(2)=6(2)^2-2(2)=20\).

a) \(f'(x)=3x^5-12x^2+2\) b) \(f'(x)=2x^2-x\) c) \(f'(x)=6x^2-2x\); the slope at \(x=2\) is \(20\).

- The derivative gives the slope at a point. - A tangent line requires both a point and a slope. - The tangent and the curve have the same slope at the point of tangency.

1. Differentiate: \(f'(x)=4x-8\). 2. Evaluate the derivative: \(f'(3)=4(3)-8=4\). 3. Find the point of tangency: \(f(3)=2(3)^2-8(3)+5=-1\), so the point is \((3,-1)\). 4. Use point-slope form: \(y+1=4(x-3)\), so \(y=4x-13\).

a) \(4\) b) \(y=4x-13\)

- Use the power rule and constant-multiple rule. - What is the derivative of a term with no variable? - You can distribute division across the terms in a numerator. - Differentiate with respect to the variable used in each function.

1. Differentiate the first function term by term: \(y'=x^3-4x\). 2. The derivative of \(15\) is \(0\), so \(y'=-6\). 3. Differentiate with respect to \(t\): \(z'=12t^2-10t\). 4. Rewrite \(f(x)=\frac{2}{3}x^3-3x\). Then \(f'(x)=2x^2-3\). 5. Treat \(a\), \(b\), and \(c\) as constants: \(y'=2ax+b\).

1) \(y'=x^3-4x\) 2) \(y'=-6\) 3) \(z'=12t^2-10t\) 4) \(f'(x)=2x^2-3\) 5) \(y'=2ax+b\)

- Rewrite the fifth root as a power with a rational exponent. - Apply the power rule to each term. - It is often easier to put every term in the form \(ax^r\) before differentiating.

1. Rewrite \(\sqrt[5]{x^3}=x^{\frac{3}{5}}\), so \(f(x)=4x^{\frac{3}{5}}-1.2x^{-\frac{2}{5}}\). 2. Apply the power rule: \(f'(x)=4\left(\frac{3}{5}\right)x^{-\frac{2}{5}}-1.2\left(-\frac{2}{5}\right)x^{-\frac{7}{5}}=2.4x^{-\frac{2}{5}}+0.48x^{-\frac{7}{5}}\).

\(f'(x)=2.4x^{-\frac{2}{5}}+0.48x^{-\frac{7}{5}}\)

- Use the quotient of powers property to simplify the function first. - Rewrite \(\frac{x^a}{x^b}\) as \(x^{a-b}\). - Apply the power rule after simplifying.

1. Simplify with exponent rules: \(f(x)=\frac{6x}{x^{\frac{2}{3}}}=6x^{\frac{1}{3}}\). 2. Apply the power rule: \(f'(x)=6\left(\frac{1}{3}\right)x^{-\frac{2}{3}}=2x^{-\frac{2}{3}}\).

\(f'(x)=2x^{-\frac{2}{3}}=\frac{2}{\sqrt[3]{x^2}}\)

- Substitute the given point into the function to find the parameter. - Differentiate the power function after finding the parameter. - A point and slope determine a line. - An x-intercept has y-coordinate \(0\). - Solve the line equation after setting \(y=0\).

1. Substitute \((2,2)\) into \(f(x)=ax^3\): \(2=8a\), so \(a=\frac{1}{4}\). 2. Differentiate: \(f'(x)=3ax^2\). With \(a=\frac{1}{4}\), \(f'(2)=3\), so the tangent slope is \(3\). 3. A line with slope \(3\) through \((2,2)\) satisfies \(y-2=3(x-2)\). Thus, \(y=3x-4\). 4. At the x-intercept, \(y=0\). Solve \(0=3x-4\), giving \(x=\frac{4}{3}\). The intercept is \((\frac{4}{3},0)\).

a) \(a=\frac{1}{4}\) b) \(f'(2)=3\), and \(y=3x-4\) c) \((\frac{4}{3},0)\)

- Use the derivative to find each tangent slope. - Find the corresponding point on the graph. - Use point-slope form for each tangent line. - The tangent slope equals the derivative at the input.

1. Differentiate: \(f'(x)=3x^2-4x\). 2. At \(x=2\), \(f(2)=0\) and \(f'(2)=4\). The tangent line is \(y=4(x-2)=4x-8\). 3. At \(x=1\), \(f(1)=-1\) and \(f'(1)=-1\). The tangent line is \(y+1=-(x-1)\), so \(y=-x\).

a) \(y=4x-8\) b) \(y=-x\)

- Write the general tangent line at \(x=a\). - The derivative gives the slope in terms of \(a\). - Evaluate the original function at \(a\). - Expand and combine like terms.

1. Differentiate: \(f'(x)=4x-5\). 2. At \(x=a\), the tangent slope is \(f'(a)=4a-5\), and the point is \((a,2a^2-5a)\). 3. Use point-slope form: \(y=(4a-5)(x-a)+2a^2-5a\). 4. Expand and simplify: \(y=(4a-5)x-4a^2+5a+2a^2-5a=(4a-5)x-2a^2\).

\(t_a: y=(4a-5)x-2a^2\)

- The derivative at \(x=4\) gives the tangent slope. - Use point-slope form for the line. - Relate slope and angle using tangent.

1. Find the point of tangency: \(f(4)=2\), so the point is \((4,2)\). 2. Differentiate: \(f'(x)=\frac{1}{4}x\). Thus, \(f'(4)=1\). 3. Write the tangent line: \(y-2=x-4\), so \(y=x-2\). 4. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

1. \(y=x-2\) 2. \(45^\circ\)

- Compare how each exponent changes and what coefficient appears. - Look for a rule connecting the original exponent to the new coefficient and exponent. - To reverse the derivative in part c), which original exponent would produce \(x^{11}\)?

1. Apply the power rule: \(f_1'(x)=3x^2\), \(f_2'(x)=5x^4\), and \(f_3'(x)=7x^6\). 2. The pattern is \(\frac{d}{dx}(x^n)=nx^{n-1}\). 3. To reverse the power rule in part c), increase the exponent from \(11\) to \(12\) and divide the coefficient by \(12\). One possible function is \(h(x)=x^{12}\). More generally, \(h(x)=x^{12}+C\).

a) \(f_1'(x)=3x^2\), \(f_2'(x)=5x^4\), and \(f_3'(x)=7x^6\) b) \(\frac{d}{dx}(x^n)=nx^{n-1}\) c) One possible function is \(h(x)=x^{12}\).

- Rewrite variables in denominators using negative exponents. - Use exponent rules to simplify products with the same base. - Apply the power rule by multiplying by the exponent and subtracting \(1\) from it.

1. For part a), the power rule gives \(f'(x)=13x^{12}\). 2. Rewrite part b) as \(g(x)=x^{-5}\). Then \(g'(x)=-5x^{-6}=-\frac{5}{x^6}\). 3. For part c), \(h'(x)=-9x^{-10}=-\frac{9}{x^{10}}\). 4. Simplify part d): \(s(t)=t^{-4}\). Then \(s'(t)=-4t^{-5}=-\frac{4}{t^5}\).

a) \(f'(x)=13x^{12}\) b) \(g'(x)=-\frac{5}{x^6}\), where \(x\neq0\) c) \(h'(x)=-\frac{9}{x^{10}}\), where \(x\neq0\) d) \(s'(t)=-\frac{4}{t^5}\), where \(t\neq0\)

- First find the complete point of tangency. - Use the power rule to find the derivative. - Evaluate the derivative at \(x=-2\). - Use point-slope form for the tangent line.

1. Find the point of tangency: \(f(-2)=\frac{1}{4}(-2)^4-(-2)^2=0\), so \(P=(-2,0)\). 2. Differentiate: \(f'(x)=x^3-2x\). 3. Find the slope: \(f'(-2)=-8+4=-4\). 4. Use point-slope form: \(y=-4(x+2)\), so \(y=-4x-8\).

\(y=-4x-8\)

- Rewrite reciprocals and radicals using negative or rational exponents. - A constant factor remains when differentiating. - Apply \(\frac{d}{dx}(x^n)=nx^{n-1}\).

1. For part a), \(f'(x)=2.5(4x^3)=10x^3\). 2. For part b), \(f'(x)=\frac{5}{12}(6x^5)=\frac{5}{2}x^5\). 3. Rewrite part c) as \(f(x)=-3x^{-2}\). Then \(f'(x)=6x^{-3}=\frac{6}{x^3}\). 4. Rewrite part d) as \(f(x)=6x^{1/2}\). Then \(f'(x)=3x^{-1/2}=\frac{3}{\sqrt{x}}\).

a) \(f'(x)=10x^3\) b) \(f'(x)=\frac{5}{2}x^5\) c) \(f'(x)=\frac{6}{x^3}\), where \(x\neq0\) d) \(f'(x)=\frac{3}{\sqrt{x}}\), where \(x>0\)

- Differentiate a sum one term at a time. - Keep constant coefficients when applying the power rule. - What is the derivative of a constant? - Recall that \(x=x^1\).

1. Differentiate \(4x^6\) to get \(24x^5\). 2. Differentiate \(-\frac{1}{2}x^4\) to get \(-2x^3\). 3. The derivative of \(3x\) is \(3\), and the derivative of \(-5\) is \(0\). 4. Therefore, \(f'(x)=24x^5-2x^3+3\).

\(f'(x)=24x^5-2x^3+3\)

- Track how each exponent and coefficient changes under the power rule. - A term with no \(x\) has derivative \(0\). - Differentiate the function one term at a time. - Keep constant coefficients, including radicals, when differentiating.

1. Differentiate term by term using the power and constant-multiple rules. 2. The derivative is \(f'(x)=5x^5-6x^4+2\sqrt{7}x\).

\(f'(x)=5x^5-6x^4+2\sqrt{7}x\)

- Radicals and \(\pi\) are constants with respect to \(x\). - What is the derivative of a term of the form \(cx\)? - Differentiate a sum term by term. - Decrease each power exponent by exactly \(1\).

1. Differentiate each term using the power, constant-multiple, and sum rules. 2. The polynomial terms give \(2x^7-4x^5+3x^3\). 3. The derivative of \(-\sqrt[5]{10}x\) is \(-\sqrt[5]{10}\), and the derivative of \(\pi\) is \(0\). 4. Therefore, \(f'(x)=2x^7-4x^5+3x^3-\sqrt[5]{10}\).

\(f'(x)=2x^7-4x^5+3x^3-\sqrt[5]{10}\)

- Convert the angle to a slope using tangent. - Set the derivative equal to that slope. - Find the y-coordinate after solving for the x-coordinate. - A point requires both coordinates.

1. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-5\). 3. Set the derivative equal to the required slope: \(2x-5=-1\). Thus, \(x=2\). 4. Evaluate the function: \(f(2)=4-10+2=-4\). Therefore, the point is \((2,-4)\).

\((2,-4)\)

- Expand products into sums of powers before differentiating. - Distribute a negative sign carefully across parentheses. - Use binomial identities when helpful. - After simplifying, differentiate each term with the power rule.

1. Expand part a): \(f(x)=4x^3-3x^2+20x-15\). Then \(f'(x)=12x^2-6x+20\). 2. Simplify part b): \(h(t)=2t^2+8t-(t^2-2t+1)=t^2+10t-1\). Then \(h'(t)=2t+10\). 3. Simplify part c): \(g(z)=\frac{1}{5}z^3-2z+z^2-4\). Then \(g'(z)=\frac{3}{5}z^2+2z-2\).

a) \(f'(x)=12x^2-6x+20\) b) \(h'(t)=2t+10\) c) \(g'(z)=\frac{3}{5}z^2+2z-2\)

- Treat parameters as constants. - Use a difference-of-squares identity in part b). - Expand the squared binomial in part c) before multiplying. - Distinguish the variable from constant factors in each term.

1. Expand part a): \(f(x)=kx^2+2akx+ka^2-bx\). Therefore, \(f'(x)=2kx+2ak-b\). 2. Use the difference of squares in part b): \((t^2+4)(t^2-4)=t^4-16\). Thus \(s(t)=t^4+\frac{2}{3}t^3-16\), and \(s'(t)=4t^3+2t^2\). 3. Expand part c): \(q(r)=3r^3-12r^2+12r\). Therefore, \(q'(r)=9r^2-24r+12\).

a) \(f'(x)=2kx+2ak-b\) b) \(s'(t)=4t^3+2t^2\) c) \(q'(r)=9r^2-24r+12\)

- Rewrite reciprocals and radicals as powers before differentiating. - A constant term has derivative \(0\). - Differentiate a sum term by term. - Treat parameters as constants with respect to \(x\).

1. Differentiate each term in part a): \(f'(x)=0.2\cdot5x^4-2\cdot3x^2+4=x^4-6x^2+4\). 2. For part b), rewrite \(f(x)=4x^{-1}+6x^{1/2}\). Then \(f'(x)=-4x^{-2}+3x^{-1/2}=-\frac{4}{x^2}+\frac{3}{\sqrt{x}}\), for \(x>0\). 3. For part c), treat \(a\), \(c\), and \(n\) as constants. The power rule gives \(f'(x)=ncx^{n-1}\).

a) \(f'(x)=x^4-6x^2+4\) b) \(f'(x)=-\frac{4}{x^2}+\frac{3}{\sqrt{x}}\), for \(x>0\) c) \(f'(x)=ncx^{n-1}\)

- A tangent line requires a point and a slope. - Use the derivative to find the slope at \(x=1\). - Evaluate the original function to find the point. - Use point-slope form.

1. Find the point of tangency: \(f(1)=0.25-2+5=3.25\), so the point is \((1,3.25)\). 2. Differentiate: \(f'(x)=x^3-4x\). 3. Find the slope: \(f'(1)=1-4=-3\). 4. Use point-slope form: \(y-3.25=-3(x-1)\), so \(y=-3x+6.25\).

\(y=-3x+6.25\)

- How are a line's slope and its angle of inclination related? - First find the tangent slope at the given input. - Use an inverse trigonometric function.

1. Differentiate: \(f'(x)=x\). 2. The tangent slope at \(x=1\) is \(m=f'(1)=1\). 3. Since \(m=\tan(\alpha)\), \(\alpha=\arctan(1)=45^\circ\).

\(\alpha=45^\circ\)

- A line with negative slope has an angle of inclination between \(90^\circ\) and \(180^\circ\). - Convert a negative inverse-tangent result to the standard inclination interval by adding \(180^\circ\).

1. Differentiate: \(f'(x)=-2x\). 2. At \(x=0.5\), the slope is \(m=f'(0.5)=-1\). 3. The calculator value \(\arctan(-1)=-45^\circ\) is coterminal with an angle of inclination of \(135^\circ\) in the interval \([0^\circ,180^\circ)\).

\(135^\circ\)

- A horizontal tangent has slope \(0\). - Rewrite \(\frac{1}{x}\) using a negative exponent before differentiating. - Use the original function to find the y-coordinate.

1. Rewrite \(f(x)=x^{-1}+x\). By the power rule, \(f^{\prime}(x)=-x^{-2}+1=-\frac{1}{x^2}+1\). 2. Evaluate the derivative: \(f^{\prime}(1)=-1+1=0\), so the tangent is horizontal. 3. Evaluate the original function: \(f(1)=1+1=2\). The point of tangency is \((1, 2)\).

\(f^{\prime}(x)=-\frac{1}{x^2}+1\), \(f^{\prime}(1)=0\), and the point of tangency is \((1, 2)\).

- Evaluate the function to find the point on the graph. - The derivative value is the tangent slope. - Use point-slope form or slope-intercept form.

1. Find the point of tangency: \(f(3)=0.5\cdot3^2-2\cdot3+1=-0.5\), so the point is \((3,-0.5)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(f'(3)=1\). 3. Use point-slope form: \(y+0.5=x-3\), so \(y=x-3.5\).

\(t: y=x-3.5\)

- How is the slope of a tangent line related to the derivative? - When are two lines parallel? - What information do you need to write an equation of a line? - Recall the tangent-line formula at \(x=a\).

1. Rewrite \(f(x) = 8x^{-1}\) and differentiate: \(f'(x) = -8x^{-2} = -\frac{8}{x^2}\). 2. At \(x = 2\), the tangent slope is \(f'(2) = -\frac{8}{2^2} = -2\). 3. The slope of \(g\) is also \(-2\), so the tangent line and \(g\) are parallel. 4. Find the point of tangency: \(f(2) = \frac{8}{2} = 4\), so \(P=(2, 4)\). 5. Use point-slope form: \(y - 4 = -2(x - 2)\). Therefore, \(t: y = -2x + 8\).

a) Since \(f'(2) = -2\), the tangent line has the same slope as \(g\), so the lines are parallel. b) \(t: y = -2x + 8\)

- What must be true about the slopes of parallel lines? - Read the slope of the given line from its equation. - Use the slope and point of tangency to write the tangent line equation.

1. Differentiate: \(g'(x)=x-3\). 2. Parallel lines have equal slopes. The given line has slope \(2\), so solve \(g'(b)=2\): \(b-3=2\), giving \(b=5\). 3. Evaluate the function: \(g(5)=\frac{1}{2}\cdot 5^2-3\cdot 5=-2.5\). Thus, \(B=(5,-2.5)\). 4. Use slope \(2\) and point \((5,-2.5)\): \(y+2.5=2(x-5)\). 5. Simplify: \(y=2x-12.5\).

a) \(g'(x)=x-3\) b) \(B=(5,-2.5)\) c) \(y=2x-12.5\)

- Evaluate the derivative to find a slope at a specified input. - Set the derivative equal to the required slope and solve. - Parallel lines have equal slopes. - To find a point on the graph, substitute its x-coordinate into the original function.

1. Evaluate the derivative: \(f'(-1)=3(-1)^2-12=-9\) and \(f'(3)=3(3)^2-12=15\). 2. For slope \(15\), solve \(3x^2-12=15\). Then \(x^2=9\), so \(x=-3\) or \(x=3\). 3. A line parallel to \(y=-9x+2\) must have slope \(-9\). Solve \(3x^2-12=-9\), which gives \(x^2=1\), so \(x=-1\) or \(x=1\). 4. Evaluate the function: \(f(-1)=16\) and \(f(1)=-6\). The points are \((-1,16)\) and \((1,-6)\). 5. At \(x=0\), \(f(0)=5\) and \(f'(0)=-12\). The tangent line is \(y-5=-12(x-0)\), or \(y=-12x+5\).

a) \(f'(-1)=-9\) and \(f'(3)=15\) b) \(x=-3\) and \(x=3\) c) \((-1,16)\) and \((1,-6)\) d) \(y=-12x+5\)

- Evaluate the derivative to find the slope at a specified input. - Set the derivative equal to the required slope and solve the resulting equation. - Read the slope of a line from its equation. - Substitute each x-coordinate into the original function to find the corresponding point.

1. Evaluate the derivative: \(g'(-2)=(-2)^2+2(-2)-3=-3\) and \(g'(1)=1^2+2(1)-3=0\). 2. For slope \(5\), solve \(x^2+2x-3=5\). Factoring \(x^2+2x-8=(x+4)(x-2)\) gives \(x=-4\) or \(x=2\). 3. A tangent parallel to \(y=-3x+10\) must have slope \(-3\). Solve \(x^2+2x-3=-3\), or \(x(x+2)=0\). Thus, \(x=0\) or \(x=-2\). 4. Evaluate the function: \(g(0)=0\) and \(g(-2)=\frac{22}{3}\). Therefore, the points are \((0,0)\) and \((-2,\frac{22}{3})\).

a) \(g'(-2)=-3\) and \(g'(1)=0\) b) \(x=-4\) and \(x=2\) c) \((0,0)\) and \((-2,\frac{22}{3})\)

- Find the point of tangency and the derivative at \(x=3\). - A line is determined by a point and its slope. - The derivative gives the tangent line's slope. - What is the y-coordinate of every point on the x-axis?

1. Find the point of tangency: \(f(3)=\frac{1}{2}\cdot 3^2-2\cdot 3+1=-0.5\), so \(P=(3,-0.5)\). 2. Differentiate: \(f'(x)=x-2\). 3. Find the tangent slope: \(f'(3)=1\). 4. Write the tangent line: \(y+0.5=x-3\), so \(y=x-3.5\). 5. At an x-intercept, \(y=0\). Solve \(0=x-3.5\) to get \(x=3.5\).

\((3.5,0)\)

- First find the point where the tangent touches the graph. - Use the derivative to find the tangent slope. - Write the tangent line equation. - A point on the x-axis has y-coordinate \(0\).

1. Find the point of tangency: \(g(1)=1^3-4(1)=-3\), so the point is \((1,-3)\). 2. Differentiate: \(g'(x)=3x^2-4\). 3. Find the tangent slope: \(g'(1)=3(1)^2-4=-1\). 4. Write the tangent line: \(y+3=-(x-1)\), so \(y=-x-2\). 5. Set \(y=0\) to find the x-intercept: \(0=-x-2\), giving \(x=-2\).

\((-2,0)\)

- Find the slope of a line whose angle of inclination is \(135^\circ\). - Express the tangent slope at an unknown input using the derivative. - Set the derivative equal to the required slope and solve.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=\frac{1}{2}x+1\). 3. Set the derivative equal to the required slope: \(\frac{1}{2}x_0+1=-1\). 4. Solve: \(\frac{1}{2}x_0=-2\), so \(x_0=-4\).

\(x_0=-4\)

- Convert the given angle to a slope. - Set the derivative equal to that slope to find the x-coordinate. - Substitute the x-coordinate into the original function, then write the tangent line.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-4\). 3. Set the derivative equal to the tangent slope: \(2x-4=-1\). Solving gives \(x=1.5\). 4. Find the y-coordinate: \(f(1.5)=(1.5)^2-4(1.5)+2=-1.75\). Thus, \(B=(1.5,-1.75)\). 5. Use point-slope form: \(y+1.75=-(x-1.5)\), so \(y=-x-0.25\).

\(B=(1.5,-1.75)\) and \(y=-x-0.25\)

- Evaluate the function to find the point of tangency. - Use the derivative for the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Use inverse tangent to find each angle of inclination.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4=0\), so the point is \((4,0)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(m_t=f'(4)=2\). 3. The tangent line is \(y=2(x-4)=2x-8\). 4. The normal slope is \(m_n=-\frac{1}{2}\), so the normal line is \(y=-0.5(x-4)=-0.5x+2\). 5. The tangent angle is \(\alpha_t=\arctan(2)\approx63.43^\circ\). The normal angle is \(\alpha_n=\arctan(-0.5)+180^\circ\approx153.43^\circ\).

\(t: y=2x-8\) \(n: y=-0.5x+2\) \(\alpha_t\approx63.43^\circ\), \(\alpha_n\approx153.43^\circ\)

- What relationship must the slopes of two perpendicular nonvertical lines satisfy? - Use the derivative to express the tangent slopes at the two intercepts. - Substitute those slopes into the perpendicularity condition. - Use the condition \(k>0\) when choosing the solution.

1. Differentiate: \(f_k'(x)=2kx\). 2. The tangent slopes are \(m_P=f_k'(-2)=-4k\) and \(m_Q=f_k'(2)=4k\). 3. A right angle at \(S\) means the two tangent lines are perpendicular, so their slopes satisfy \(m_Pm_Q=-1\). 4. Substitute: \((-4k)(4k)=-1\), so \(-16k^2=-1\) and \(k^2=\frac{1}{16}\). 5. Since \(k>0\), \(k=\frac{1}{4}\).

\(k=\frac{1}{4}\)

- The tangent slope is the derivative at the point. - The slopes of a tangent and its normal are negative reciprocals when both are defined. - Convert an angle of inclination to a slope using tangent. - To verify a tangent line, check both the point and the slope.

1. Differentiate: \(f'(x)=x-3\). 2. At \(x=4\), \(f(4)=-3\) and \(f'(4)=1\). The tangent line is \(y+3=x-4\), or \(y=x-7\). The normal slope is \(-1\), so the normal line is \(y+3=-(x-4)\), or \(y=-x+1\). 3. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). Solve \(x-3=-1\) to get \(x=2\). 4. At \(x=2\), \(f(2)=-3\) and \(f'(2)=-1\). The tangent line is \(y+3=-(x-2)\), which simplifies to \(y=-x-1\). Therefore, the given line is tangent at \(x=2\).

a) Tangent: \(y=x-7\); normal: \(y=-x+1\) b) \(x=2\) c) Yes. At \(x=2\), the point is \((2,-3)\) and the tangent slope is \(-1\), giving \(y=-x-1\).

- A line parallel to the x-axis has slope \(0\). - Substitute the given input into the original function to find the point. - A normal line has slope equal to the negative reciprocal of the tangent slope. - Use point-slope form to write each line.

1. Differentiate: \(f'(x)=x^2-1\). 2. At \(x=2\), \(f(2)=\frac{2}{3}\) and \(f'(2)=3\). The tangent line is \(y-\frac{2}{3}=3(x-2)\), or \(y=3x-\frac{16}{3}\). 3. A tangent parallel to the x-axis has slope \(0\). Solve \(x^2-1=0\), giving \(x=-1\) and \(x=1\). 4. At \(x=0\), \(f(0)=0\) and the tangent slope is \(-1\). The normal slope is \(1\), so the normal line is \(y=x\).

a) \(y=3x-\frac{16}{3}\) b) \(x=-1\) and \(x=1\) c) \(y=x\)

- The derivative gives the graph's slope at a point. - A tangent line with slope \(0\) is horizontal. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form for each line.

1. Differentiate: \(f'(x)=x^3-4x\). 2. At \(x=1\), \(f'(1)=1-4=-3\). 3. At \(x=2\), \(f(2)=1\) and \(f'(2)=0\). The tangent line is horizontal, so its equation is \(y=1\). 4. At \(x=-1\), \(f(-1)=\frac{13}{4}\) and the tangent slope is \(f'(-1)=3\). The normal slope is \(-\frac{1}{3}\). 5. The normal line is \(y-\frac{13}{4}=-\frac{1}{3}(x+1)\), which simplifies to \(y=-\frac{1}{3}x+\frac{35}{12}\).

a) \(-3\) b) \(y=1\) c) \(y=-\frac{1}{3}x+\frac{35}{12}\)

- Convert the known normal slope to the tangent slope. - The derivative gives the tangent slope at each input. - Set the derivative equal to the required tangent slope. - Evaluate the function before writing the line equation.

1. The tangent slope is the negative reciprocal of the normal slope: \(m_t=-\frac{1}{0.5}=-2\). 2. Differentiate: \(f'(x)=2x-4\). Set \(f'(x_0)=-2\): \(2x_0-4=-2\), so \(x_0=1\). 3. Evaluate the function: \(f(1)=2\), so the point is \((1,2)\). 4. The normal line is \(y-2=0.5(x-1)\), or \(y=0.5x+1.5\).

a) \(x_0=1\) b) \(n: y=0.5x+1.5\)

- Find the tangent slope at \(A\) using the derivative. - Convert the tangent slope to the normal slope. - Write the normal line through \(A\). - Solve the two line equations as a system.

1. Differentiate: \(p'(x)=2x-4\). The tangent slope at \(x=3\) is \(p'(3)=2\). 2. The normal slope is \(-\frac{1}{2}\). 3. Through \(A(3,3)\), the normal line is \(y-3=-0.5(x-3)\), or \(y=-0.5x+4.5\). 4. Set the normal equation equal to \(y=x-3\): \(-0.5x+4.5=x-3\). Thus, \(7.5=1.5x\), so \(x=5\). 5. Then \(y=5-3=2\), so \(S=(5,2)\).

\(S=(5,2)\)

- Set the derivative equal to the given slope. - Substitute each solution into the original function to find the points. - Use a coordinate-area formula for a triangle whose one vertex is the origin. - Take the absolute value so the area is positive.

1. Differentiate: \(f'(x)=x^2-2x-2\). 2. Set the derivative equal to \(1\): \(x^2-2x-2=1\), or \(x^2-2x-3=0\). Thus, \(x=-1\) or \(x=3\). 3. Find the corresponding points: \(f(-1)=\frac{2}{3}\) and \(f(3)=-6\). Thus, the points are \((-1,\frac{2}{3})\) and \((3,-6)\). 4. With one vertex at the origin, the area is \(\frac{1}{2}|x_1y_2-x_2y_1|\). 5. Therefore, \(A=\frac{1}{2}|(-1)(-6)-3(\frac{2}{3})|=\frac{1}{2}|6-2|=2\).

\(2\) square units

- The y-intercept occurs where \(x=0\). - A horizontal tangent has derivative \(0\). - The derivative at a point gives the tangent slope. - Use point-slope form to write each tangent line.

1. The y-intercept occurs at \(x=0\), and \(f(0)=11\). Differentiate: \(f'(x)=3x^2-6x-9\), so \(f'(0)=-9\). The tangent line is \(y-11=-9(x-0)\), or \(y=-9x+11\). 2. A horizontal tangent has slope \(0\). Solve \(3x^2-6x-9=0\). Dividing by \(3\) gives \(x^2-2x-3=0\), so \(x=-1\) or \(x=3\). 3. At \(x=4\), \(f(4)=-9\) and \(f'(4)=15\). Thus, \(P=(4,-9)\), and the tangent line is \(y+9=15(x-4)\), or \(y=15x-69\).

1. At \(x=0\), \(f(0)=11\) and \(f'(0)=-9\), so the tangent is \(y=-9x+11\). 2. \(x=-1\) and \(x=3\) 3. \(P=(4,-9)\) and \(y=15x-69\)

- Find the point of tangency from the original function. - Use the derivative to find the tangent slope. - Write the line using point-slope form. - Relate the slope to the angle using tangent.

1. Find the point of tangency: \(f(2)=2^3-2(2)^2+1=1\), so \(P=(2,1)\). 2. Differentiate: \(f'(x)=3x^2-4x\). 3. Find the slope: \(f'(2)=3(2)^2-4(2)=4\). 4. Write the tangent line: \(y-1=4(x-2)\), so \(y=4x-7\). 5. The angle \(\alpha\) satisfies \(\tan(\alpha)=4\), so \(\alpha=\arctan(4)\approx 75.96^\circ\).

Tangent line: \(y=4x-7\) Angle: \(75.96^\circ\), approximately

- Use the derivative to find the tangent slope. - Substitute \(x=1\) into the original function to find the point. - Relate the slope to the angle using tangent. - Write the line in slope-intercept or point-slope form.

1. Find the point of tangency: \(f(1)=-0.25(1)^4+1^2+1=1.75\), so \(P=(1,1.75)\). 2. Differentiate: \(f'(x)=-x^3+2x\). 3. Find the slope: \(f'(1)=1\). 4. Write the tangent line: \(y-1.75=x-1\), so \(y=x+0.75\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

Tangent line: \(y=x+0.75\) Angle: \(45^\circ\)

- Parallel lines have equal slopes. - Set the derivative equal to the specified slope. - Find the corresponding points on the graph. - Use each point and the common slope to write a tangent line.

1. Differentiate: \(f'(x)=x^2-4\). 2. The given line has slope \(5\). Set \(f'(x)=5\): \(x^2-4=5\), so \(x=-3\) or \(x=3\). 3. Evaluate the function: \(f(-3)=5\) and \(f(3)=-1\). 4. At \((-3,5)\), the tangent line is \(y-5=5(x+3)\), or \(y=5x+20\). 5. At \((3,-1)\), the tangent line is \(y+1=5(x-3)\), or \(y=5x-16\).

\(y=5x+20\) and \(y=5x-16\)

- Treat the parameter as a constant when differentiating. - Set the derivative at \(x=2\) equal to the required slope. - Find the point of tangency using the corresponding function. - Use point-slope form for the tangent line.

1. Differentiate while treating \(k\) as a constant: \(g_k'(x)=3x^2-2kx\). 2. Require slope \(4\) at \(x=2\): \(3(2)^2-2k(2)=4\). Thus, \(12-4k=4\), so \(k=2\). 3. For \(k=2\), \(g_2(2)=2^3-2(2)^2+4=4\). The point of tangency is \((2,4)\). 4. Using slope \(4\), the tangent line is \(y-4=4(x-2)\), or \(y=4x-4\).

a) \(g_k'(x)=3x^2-2kx\) b) \(k=2\) c) \(y=4x-4\)

- Use the derivative to find the tangent slope. - Find the complete point of tangency. - Use point-slope form to write the line. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(f'(x)=x-2\). 2. At \(x=-1\), \(f(-1)=1.5\) and \(f'(-1)=-3\). 3. Write the tangent line: \(y-1.5=-3(x+1)\), so \(y=-3x-1.5\). 4. At the y-intercept, \(x=0\). Thus, \(y=-1.5\), and the intercept is \((0,-1.5)\).

Tangent line: \(y=-3x-1.5\) Y-intercept: \((0,-1.5)\)

- Evaluate the function to find the full point. - Use the derivative for the tangent slope. - Perpendicular line slopes are negative reciprocals. - Use point-slope form for both lines.

1. Differentiate: \(g'(x)=x^2-2x\). 2. Evaluate the function and derivative: \(g(3)=2\) and \(g'(3)=3\). 3. The tangent line is \(y-2=3(x-3)\), or \(y=3x-7\). 4. The normal slope is \(-\frac{1}{3}\). Thus, the normal line is \(y-2=-\frac{1}{3}(x-3)\), or \(y=-\frac{1}{3}x+3\).

a) \(t: y=3x-7\) b) \(n: y=-\frac{1}{3}x+3\)

- What value must the derivative have at a horizontal tangent? - Treat \(a\) as a constant while differentiating with respect to \(x\). - Substitute the given x-coordinate into the derivative equation.

1. Treat \(a\) as a constant and apply the power rule: \(g_a'(x)=3ax^2-8x+a\). 2. A horizontal tangent at \(x=1\) requires \(g_a'(1)=0\). 3. Substitute \(x=1\): \(3a-8+a=0\). 4. Solve \(4a-8=0\) to obtain \(a=2\).

a) \(g_a'(x)=3ax^2-8x+a\) b) \(a=2\)

- Substitute a point into the original function to find a parameter. - The derivative gives the slope at a specified x-value. - Apply the power rule term by term. - The origin has x-coordinate \(0\).

1. Substitute \((2, 8)\): \(8-4k+4=8\). Solving gives \(k=1\). 2. Differentiate: \(f_k'(x)=3x^2-2kx+2\). 3. At the origin, \(f_k'(0)=2\), independent of \(k\). 4. For slope \(11\) at \(x=3\), solve \(f_k'(3)=11\): \(27-6k+2=11\). 5. This gives \(6k=18\), so \(k=3\).

a) \(k=1\) b) The slope is \(2\). c) \(k=3\)

- Substitute the given point into the function. - Differentiate to find a slope formula. - Evaluate the derivative at the specified x-value. - Decide whether the requested slope is constant or depends on the parameter.

1. Substitute \((1, 2)\): \(1-3+t=2\), so \(t=4\). 2. Differentiate: \(g_t'(x)=3x^2-6x+t\). 3. At the origin, \(g_t'(0)=t\). 4. For slope \(5\) at \(x=2\), solve \(g_t'(2)=5\): \(12-12+t=5\). 5. Therefore, \(t=5\).

a) \(t=4\) b) The slope is \(t\). c) \(t=5\)

- Convert each angle to a slope using tangent. - Set the derivative equal to the required slope. - Check whether the resulting squared equation has real solutions. - Consider the range of \(g'(x)=x^2-1\).

1. An angle of \(45^\circ\) corresponds to slope \(\tan(45^\circ)=1\). Differentiate: \(g'(x)=x^2-1\). 2. Solve \(x^2-1=1\). Then \(x^2=2\), so \(x=\pm\sqrt{2}\). 3. Evaluate the function: \(g(\sqrt{2})=-\frac{\sqrt{2}}{3}\) and \(g(-\sqrt{2})=\frac{\sqrt{2}}{3}\). The points are \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\). 4. An angle of \(120^\circ\) corresponds to slope \(\tan(120^\circ)=-\sqrt{3}\). The equation \(x^2-1=-\sqrt{3}\) becomes \(x^2=1-\sqrt{3}\). 5. Since \(1-\sqrt{3}<0\), there is no real solution. Equivalently, \(g'(x)=x^2-1\ge -1\), so it can never equal \(-\sqrt{3}\).

1. \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\) 2. No such point exists because \(g'(x)\ge -1\), while \(\tan(120^\circ)=-\sqrt{3}<-1\).

- A horizontal tangent has slope \(0\). - Evaluate the derivative at both specified inputs. - Compare absolute values when comparing steepness.

1. Differentiate: \(g'(x)=x^2-4x+3\). 2. Horizontal tangent lines occur where \(g'(x)=0\). Factor \(x^2-4x+3=(x-1)(x-3)\), so \(x=1\) and \(x=3\). 3. Evaluate the slopes: \(g'(0)=3\) and \(g'(4)=3\). 4. Steepness is determined by the absolute value of slope. Since both slopes have magnitude \(3\), the graph is equally steep at the two points.

a) \(x=1\) and \(x=3\) b) The graph is equally steep at \(x=0\) and \(x=4\); the slope is \(3\) at each point.

- Find the point and the derivative at \(x=1\) in terms of \(k\). - Write the tangent line in point-slope form. - Substitute the coordinates of \(Q\) into the tangent-line equation. - Solve the resulting equation for \(k\).

1. At \(x=1\), the point on the graph is \(B(1, f_k(1))=(1, 5-k)\). 2. Differentiate: \(f_k'(x)=4x^3-2kx\), so the tangent slope at \(x=1\) is \(4-2k\). 3. The tangent line is \(y-(5-k)=(4-2k)(x-1)\). 4. Because the line passes through \(Q(0, 1)\), \(1-(5-k)=(4-2k)(-1)\). 5. Simplifying gives \(k-4=-4+2k\), so \(k=0\).

\(k=0\)

- Simplify the rational expression before differentiating. - The derivative value gives the tangent slope. - Use point-slope form for the tangent line. - An even power in the denominator has the same sign on both sides of its zero.

1. Simplify before differentiating: \(f(x)=\frac{9x^2-12x+4}{2x}=4.5x-6+\frac{2}{x}\). 2. Differentiate: \(f'(x)=4.5-\frac{2}{x^2}=\frac{9x^2-4}{2x^2}\). 3. At \(x=1\), the slope is \(f'(1)=2.5\). Using point-slope form through \((1,0.5)\), \(y-0.5=2.5(x-1)\), so \(y=2.5x-2\). 4. The derivative is undefined at \(x=0\), while its numerator approaches \(-4\neq0\). Thus \(x=0\) is a vertical asymptote of \(f'\). Because the denominator contains \(x^2\), it is positive on both sides, so \(f'\) approaches \(-\infty\) from both sides and does not change sign across the asymptote.

a) \(y=2.5x-2\). b) The vertical asymptote of \(f'\) is \(x=0\). The derivative is negative on both sides near the asymptote, so there is no sign change.

- Recall the product rule: \((uv)'=u'v+uv'\). - Use the exponent rules \(x^a\cdot x^b=x^{a+b}\) and \(\frac{x^a}{x^b}=x^{a-b}\). - Compare the coefficient in each derivative with the original exponent. - Check whether the new exponent is exactly \(1\) less than the original exponent.

1. Using \(g(x)=x\cdot x^{\frac{1}{4}}\), the product rule gives \(g'(x)=x^{\frac{1}{4}}+x\cdot\frac{1}{4}x^{-\frac{3}{4}}=x^{\frac{1}{4}}+\frac{1}{4}x^{\frac{1}{4}}=\frac{5}{4}x^{\frac{1}{4}}\). 2. Using \(k(x)=\frac{1}{f(x)}\), \(k'(x)=\frac{-f'(x)}{(f(x))^2}=\frac{-\frac{1}{4}x^{-\frac{3}{4}}}{x^{\frac{1}{2}}}=-\frac{1}{4}x^{-\frac{5}{4}}\). 3. Both results have the form \((x^r)'=rx^{r-1}\). For a real-valued power function, the rule applies at points in its domain where the function is differentiable. In particular, it is valid for every rational \(r\) when \(x>0\).

1. \(g'(x)=\frac{5}{4}x^{\frac{1}{4}}\) 2. \(k'(x)=-\frac{1}{4}x^{-\frac{5}{4}}\) 3. \((x^r)'=rx^{r-1}\) wherever the real-valued function \(x^r\) is differentiable; in particular, for \(x>0\).

- Rewrite each radical and reciprocal as a power with a rational exponent. - The derivative evaluated at a point gives the slope of the tangent line there. - Any power of \(1\) equals \(1\). - Watch the signs when differentiating negative exponents.

1. Rewrite the function using rational exponents: \(f(x)=\frac{3}{4}x^{\frac{3}{2}}-2x^{-\frac{1}{4}}\). 2. Apply the power rule: \(f'(x)=\frac{3}{4}\cdot\frac{3}{2}x^{\frac{1}{2}}-2\left(-\frac{1}{4}\right)x^{-\frac{5}{4}}=\frac{9}{8}x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{5}{4}}\). 3. Evaluate at \(x=1\): \(f'(1)=\frac{9}{8}+\frac{1}{2}=\frac{13}{8}=1.625\).

The tangent line has slope \(\frac{13}{8}=1.625\).

- First rewrite radicals and reciprocals in the form \(x^r\). - Apply the power rule to each term. - When differentiating a power, multiply by the exponent and subtract \(1\) from the exponent. - Substituting \(x=1\) simplifies every power of \(1\) to \(1\).

1. a) By the power rule, \(f'(x)=1.2\left(\frac{1}{2}\right)x^{-\frac{1}{2}}=0.6x^{-\frac{1}{2}}\). Therefore, \(f'(1)=0.6\). 2. b) Rewrite \(f(x)=x^{\frac{7}{3}}\). Then \(f'(x)=\frac{7}{3}x^{\frac{4}{3}}\), so \(f'(1)=\frac{7}{3}\approx2.333\). 3. c) Rewrite \(f(x)=6x^{-\frac{3}{2}}\). Then \(f'(x)=-9x^{-\frac{5}{2}}\), so \(f'(1)=-9\).

a) \(f'(x)=0.6x^{-\frac{1}{2}}\); \(f'(1)=0.6\) b) \(f'(x)=\frac{7}{3}x^{\frac{4}{3}}\); \(f'(1)=\frac{7}{3}\approx2.333\) c) \(f'(x)=-9x^{-\frac{5}{2}}\); \(f'(1)=-9\)

- Simplify each expression with exponent rules before differentiating. - Recall that \(\sqrt[n]{x^m}=x^{\frac{m}{n}}\). - A power in the denominator can be written with a negative exponent. - When multiplying powers with the same base, add their exponents.

1. a) Combine the powers: \(g(x)=x^{\frac{7}{5}}x^{\frac{4}{5}}=x^{\frac{11}{5}}\). Thus, \(g'(x)=\frac{11}{5}x^{\frac{6}{5}}\), and \(g'(1)=\frac{11}{5}=2.2\). 2. b) Rewrite \(g(x)=\frac{5}{2}x^{-\frac{3}{4}}\). Thus, \(g'(x)=-\frac{15}{8}x^{-\frac{7}{4}}\), and \(g'(1)=-\frac{15}{8}=-1.875\). 3. c) Rewrite \(g(x)=x^{\frac{2}{3}-2}=x^{-\frac{4}{3}}\). Thus, \(g'(x)=-\frac{4}{3}x^{-\frac{7}{3}}\), and \(g'(1)=-\frac{4}{3}\approx-1.333\).

a) \(g'(x)=\frac{11}{5}x^{\frac{6}{5}}\); \(g'(1)=\frac{11}{5}=2.2\) b) \(g'(x)=-\frac{15}{8}x^{-\frac{7}{4}}\); \(g'(1)=-\frac{15}{8}=-1.875\) c) \(g'(x)=-\frac{4}{3}x^{-\frac{7}{3}}\); \(g'(1)=-\frac{4}{3}\approx-1.333\)

- Use the fact that \(P\) lies on the graph. - The point \(P(2, 0)\) is also the tangent line's x-intercept. - Use the triangle's area to find the tangent line's y-intercept. - Find the tangent slope from the intercepts and match it to the derivative at \(x=2\).

1. Since \(P(2, 0)\) lies on the graph, \(f_{a,b}(2)=0\). Therefore, \(4a+2b+5=0\), or \(4a+2b=-5\). 2. The tangent line crosses the x-axis at \((2, 0)\). Let its y-intercept be \((0, y_0)\). The triangle's area is \(\frac{1}{2}\cdot2\cdot y_0=6\), so \(y_0=6\). 3. The tangent line through \((0, 6)\) and \((2, 0)\) has slope \(\frac{0-6}{2-0}=-3\). 4. Differentiate: \(f_{a,b}'(x)=2ax+b\). At \(x=2\), the tangent slope gives \(4a+b=-3\). 5. Solve the system \(4a+2b=-5\) and \(4a+b=-3\). Subtraction gives \(b=-2\), and then \(a=-\frac{1}{4}\).

\(a=-\frac{1}{4}\) and \(b=-2\)

- Rewrite the quadratic in vertex form. - Substitute the point's x-coordinate into the function. - Use the derivative to verify the tangent slope at \(P\). - Use the point and slope to write the tangent line. - Test the origin by substituting \((0,0)\) into the line equation.

1. Complete the square: \(f(x)=(x-2)^2+1\). Therefore, the vertex is \((2,1)\). 2. Evaluate the function: \(f(3)=9-12+5=2\). Thus, \(P=(3,2)\) lies on the graph. 3. Differentiate: \(f'(x)=2x-4\). Then \(f'(3)=2\), so the tangent slope at \(P\) is \(2\). 4. A line with slope \(2\) through \((3,2)\) satisfies \(y-2=2(x-3)\), so \(y=2x-4\). 5. At \(x=0\), the line has value \(-4\), not \(0\). Therefore, it does not pass through the origin.

a) \((2,1)\) b) Yes, because \(f(3)=2\). c) \(f'(3)=2\), so the tangent line is \(y=2x-4\); it does not pass through the origin.

- Evaluate the function to find the full point. - The derivative value gives the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form.

1. Find the point: \(f(2)=\frac{1}{2}\cdot2^2+2=4\), so \(P=(2,4)\). 2. Differentiate: \(f'(x)=x\). The tangent slope is \(f'(2)=2\). 3. The tangent line is \(y-4=2(x-2)\), or \(y=2x\). 4. The normal slope is \(-\frac{1}{2}\). Thus, \(y-4=-\frac{1}{2}(x-2)\), or \(y=-\frac{1}{2}x+5\).

\(t: y=2x\) \(n: y=-\frac{1}{2}x+5\)

- The derivative gives the tangent slope. - A horizontal tangent has slope \(0\). - Write the general tangent line at \(x=u\). - Expand and combine like terms carefully.

1. Differentiate: \(f'(x)=3x^2-3\). 2. At \(x=2\), \(f(2)=2\) and \(f'(2)=9\). Thus, \(y-2=9(x-2)\), so \(y=9x-16\). 3. A horizontal tangent has slope \(0\). Solve \(3x^2-3=0\), giving \(x=-1\) and \(x=1\). 4. At an arbitrary input \(u\), the point is \((u,u^3-3u)\) and the slope is \(3u^2-3\). 5. Substitute into point-slope form: \(y=(3u^2-3)(x-u)+u^3-3u\). Expanding gives \(y=(3u^2-3)x-2u^3\).

a) \(y=9x-16\) b) \(x=-1\) and \(x=1\) c) Substituting \(f(u)=u^3-3u\) and \(f'(u)=3u^2-3\) into point-slope form simplifies to \(y=(3u^2-3)x-2u^3\).

- The normal slope is the negative reciprocal of the tangent slope. - Find both axis intercepts of the tangent line. - The axes and the line form a right triangle. - Use \(\frac{1}{2}(\text{base})(\text{height})\).

1. Differentiate: \(f'(x)=x\). 2. At \(x=2\), \(f(2)=0\) and the tangent slope is \(2\). Thus, the tangent line is \(y=2(x-2)=2x-4\). 3. The normal slope is the negative reciprocal, \(-\frac{1}{2}\). Thus, the normal line is \(y=-\frac{1}{2}(x-2)=-\frac{1}{2}x+1\). 4. The tangent line intersects the x-axis at \((2,0)\) and the y-axis at \((0,-4)\). 5. The right triangle has leg lengths \(2\) and \(4\), so its area is \(\frac{1}{2}\cdot 2\cdot 4=4\).

a) Tangent: \(y=2x-4\); normal: \(y=-\frac{1}{2}x+1\) b) \(4\) square units

- Rewrite the rational expression using a negative exponent. - Find \(f'(a)\) and \(f(a)\). - Use point-slope form. - Apply exponent rules carefully when simplifying.

1. Rewrite the function as \(f(x)=4x^{-2}\). 2. Differentiate using the power rule: \(f'(x)=-8x^{-3}=-\frac{8}{x^3}\). 3. At \(x=a\), the tangent slope is \(-\frac{8}{a^3}\), and the point is \((a,\frac{4}{a^2})\). 4. Use point-slope form: \(y=-\frac{8}{a^3}(x-a)+\frac{4}{a^2}\). 5. Simplify: \(y=-\frac{8}{a^3}x+\frac{8}{a^2}+\frac{4}{a^2}=-\frac{8}{a^3}x+\frac{12}{a^2}\).

\(y=-\frac{8}{a^3}x+\frac{12}{a^2}\)

- Rewrite the reciprocal term using a negative exponent. - The derivative gives the tangent slope. - Use the point and slope to write the tangent line.

1. Find the point of tangency: \(f(1)=3\), so \(P=(1,3)\). 2. Rewrite \(f(x)=2x^{-1}+x\) and differentiate: \(f'(x)=-2x^{-2}+1=-\frac{2}{x^2}+1\). 3. Find the slope: \(f'(1)=-1\). 4. Use point-slope form: \(y-3=-(x-1)\), so \(y=-x+4\).

\(y=-x+4\)

- Find both tangent slopes using the derivative. - What slope relationship characterizes perpendicular lines? - You may also compare the lines' angles of inclination.

1. Differentiate: \(g'(x)=2x-3\). 2. The tangent slopes are \(g'(1)=-1\) and \(g'(2)=1\). 3. Since the product of the slopes is \((-1)(1)=-1\), the tangent lines are perpendicular. 4. Therefore, their angle of intersection is \(90^\circ\).

\(90^\circ\)

- Parallel lines have equal slopes. - The derivative gives the tangent slope at \(a\). - Set the derivative equal to \(2\) and solve. - Check for more than one solution.

1. Differentiate: \(f'(x)=x^2+2x-6\). 2. The given line has slope \(2\), so require \(f'(a)=2\). 3. Solve \(a^2+2a-6=2\), or \(a^2+2a-8=0\). 4. Factor: \((a+4)(a-2)=0\). Thus, \(a=-4\) or \(a=2\).

\(a=-4\) and \(a=2\)

- Rewrite reciprocals and radicals using negative or rational exponents. - Use \(\frac{1}{x^n}=x^{-n}\). - Use \(\sqrt{x}=x^{1/2}\). - Check the sign when differentiating a negative exponent.

1. Rewrite the function as \(g(x)=2x^{-2}+6x^{1/2}-x^3\). 2. Differentiate: \(g'(x)=-4x^{-3}+3x^{-1/2}-3x^2\). 3. Rewrite without negative or fractional exponents: \(g'(x)=-\frac{4}{x^3}+\frac{3}{\sqrt{x}}-3x^2\).

\(g'(x)=-\frac{4}{x^3}+\frac{3}{\sqrt{x}}-3x^2\), where \(x>0\)

- The derivative represents the tangent slope. - Convert the angle to a slope. - Solve the derivative equation for all possible inputs. - Substitute each input into the original function.

1. An angle of \(45^\circ\) corresponds to slope \(1\). 2. Differentiate: \(f'(x)=x^2-x-1\). 3. Set the derivative equal to \(1\): \(x^2-x-1=1\), so \(x^2-x-2=0\). 4. Factor: \((x-2)(x+1)=0\). Thus, \(x=2\) or \(x=-1\). 5. Evaluate the function: \(f(2)=-\frac{4}{3}\) and \(f(-1)=\frac{1}{6}\). The points are \((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\).

\((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\)

- Set the function values equal to find intersection points. - Use each derivative to find the tangent slopes at an intersection. - Use the angle-between-lines formula for two known slopes. - Report the acute angle.

1. Set the functions equal: \(x^2-2x+2=-x^2+4x-2\). This simplifies to \(x^2-3x+2=0\), so \(x=1\) or \(x=2\). 2. The corresponding points are \((1,1)\) and \((2,2)\). 3. Differentiate: \(f'(x)=2x-2\) and \(g'(x)=-2x+4\). 4. At \(x=1\), the slopes are \(0\) and \(2\). Thus, \(\tan(\gamma)=\left|\frac{2-0}{1+0\cdot 2}\right|=2\), so \(\gamma=\arctan(2)\approx 63.43^\circ\). 5. At \(x=2\), the slopes are \(2\) and \(0\), giving the same value of \(\tan(\gamma)\). Therefore, the acute intersection angle is also approximately \(63.43^\circ\).

1. \((1,1)\) and \((2,2)\) 2. The acute intersection angle is approximately \(63.43^\circ\) at both points.

- Find both derivative values at each intersection. - Equal tangent slopes give an intersection angle of \(0^\circ\). - Otherwise, use the angle-between-lines formula. - Use the acute angle between the tangent lines.

1. Differentiate: \(f'(x)=3x^2-1\) and \(g'(x)=2x\). 2. At \(P=(-1,0)\), the slopes are \(2\) and \(-2\). Therefore, \(\tan(\gamma)=\left|\frac{2-(-2)}{1+2(-2)}\right|=\frac{4}{3}\), so \(\gamma=\arctan(\frac{4}{3})\approx 53.13^\circ\). 3. At \(Q=(1,0)\), both slopes equal \(2\). The tangent lines coincide, so the intersection angle is \(0^\circ\); the graphs are tangent at \(Q\).

At \(P=(-1,0)\), the acute intersection angle is approximately \(53.13^\circ\). At \(Q=(1,0)\), the angle is \(0^\circ\); the graphs are tangent there.

- Tangent and normal slopes are negative reciprocals. - A line equation requires a point and a slope. - What does a tangent slope of \(0\) imply about the normal line? - Apply the vertical line test to a vertical line.

1. At \(x=1\), \(f(1)=-2\). Also, \(f'(x)=4x\), so the tangent slope is \(f'(1)=4\). 2. The normal slope is \(-\frac{1}{4}\). Through \((1,-2)\), the normal line is \(y+2=-\frac{1}{4}(x-1)\), or \(y=-\frac{1}{4}x-\frac{7}{4}\). 3. Setting \(x=0\) gives the y-intercept \(\left(0,-\frac{7}{4}\right)\). Setting \(y=0\) gives \(x=-7\), so the x-intercept is \((-7,0)\). 4. At \(x=0\), \(f'(0)=0\), so the tangent is horizontal. The normal is the vertical line through \((0,-4)\), namely \(x=0\). 5. A vertical line has undefined slope and assigns many y-values to the same x-value, so it cannot be written as a function \(y=mx+b\).

a) \(n: y=-0.25x-1.75\) b) The intercepts are \((0,-1.75)\) and \((-7,0)\). c) The normal is the y-axis, \(x=0\). It is vertical, so its slope is undefined and it is not a function of \(x\).

- Parallel lines have equal slopes. - Express each tangent slope using a derivative. - Set the derivative expressions equal. - Solve the resulting quadratic equation.

1. Differentiate: \(f'(x)=x^2-4\) and \(g'(x)=2x-1\). 2. Parallel tangent lines have equal slopes, so solve \(x^2-4=2x-1\). 3. Rearrange: \(x^2-2x-3=0\). 4. Factor: \((x-3)(x+1)=0\). Therefore, \(x=-1\) or \(x=3\).

\(x=-1\) and \(x=3\)

- Equal slopes mean equal derivative values. - Set the derivative functions equal and solve. - For each tangent line, use its own point on the corresponding graph. - Use point-slope form.

1. Differentiate: \(f'(x)=3x^2-6x\) and \(g'(x)=6x-9\). 2. Set the slopes equal: \(3x^2-6x=6x-9\). Dividing by \(3\) gives \(x^2-4x+3=0\), so \(x=1\) or \(x=3\). 3. The smaller value is \(x=1\). The common slope is \(-3\). 4. Since \(f(1)=3\), the tangent to \(f\) is \(y-3=-3(x-1)\), or \(y=-3x+6\). 5. Since \(g(1)=-4\), the tangent to \(g\) is \(y+4=-3(x-1)\), or \(y=-3x-1\).

a) \(x=1\) and \(x=3\) b) At \(x=1\), the tangent lines are \(y=-3x+6\) and \(y=-3x-1\).

- Set the two function values equal to find their common point. - Use each derivative to find the tangent slopes at that point. - Set \(y=0\) to find each tangent line's x-intercept. - Use the x-axis segment as the base of the triangle.

1. Set the functions equal: \(x^2=(x-2)^2\). This simplifies to \(x=1\), and the common value is \(1\). Thus, \(P=(1,1)\). 2. Differentiate: \(f'(x)=2x\) and \(g'(x)=2(x-2)\). At \(x=1\), the slopes are \(2\) and \(-2\). 3. The tangent lines are \(y-1=2(x-1)\), or \(y=2x-1\), and \(y-1=-2(x-1)\), or \(y=-2x+3\). 4. Setting \(y=0\) gives x-intercepts \(x=\frac{1}{2}\) and \(x=\frac{3}{2}\). 5. The base on the x-axis has length \(1\), and the height from \(P\) is \(1\). Therefore, the area is \(\frac{1}{2}\).

a) \(P=(1,1)\) b) The tangent lines meet the x-axis at \((\frac{1}{2},0)\) and \((\frac{3}{2},0)\). c) \(\frac{1}{2}\) square unit

- Differentiate the square-root and reciprocal functions using power notation. - Use point-slope form at \((1,1)\). - The triangle's base lies on the y-axis. - The x-coordinate of \(P\) gives the height relative to that base.

1. Differentiate: \(f'(x)=\frac{1}{2\sqrt{x}}\) and \(g'(x)=-\frac{1}{x^2}\). 2. At \(x=1\), the slopes are \(\frac{1}{2}\) and \(-1\). 3. The tangent to \(f\) is \(y-1=\frac{1}{2}(x-1)\), or \(y=\frac{1}{2}x+\frac{1}{2}\). 4. The tangent to \(g\) is \(y-1=-(x-1)\), or \(y=-x+2\). 5. Their y-intercepts are \(\frac{1}{2}\) and \(2\), so the vertical base has length \(\frac{3}{2}\). The horizontal distance from \(P\) to the y-axis is \(1\). 6. The area is \(\frac{1}{2}\cdot\frac{3}{2}\cdot 1=\frac{3}{4}\).

a) \(y=\frac{1}{2}x+\frac{1}{2}\) and \(y=-x+2\) b) \(\frac{3}{4}\) square unit

- Use each derivative to express a tangent slope. - What relationship do the slopes of perpendicular lines satisfy? - Write one equation involving \(x_1\) and \(x_2\). - Solve for one input in terms of the other.

1. Differentiate: \(f'(x)=x\) and \(g'(x)=-\frac{1}{x^2}\). 2. Perpendicular nonvertical lines have slopes whose product is \(-1\). Thus, \(f'(x_1)g'(x_2)=-1\). 3. Substitute: \(x_1(-\frac{1}{x_2^2})=-1\). 4. Since \(x_2\ne 0\), this simplifies to \(x_1=x_2^2\).

\(x_1=x_2^2\), with \(x_2\ne 0\)

- Set the function values equal to find intersections. - Equal tangent slopes mean equal derivative values. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form for the normal line.

1. Set the functions equal: \(x^2+3x=-x^2+5\), so \(2x^2+3x-5=0\). The solutions are \(x=1\) and \(x=-\frac{5}{2}\). 2. The intersection points are \((1,4)\) and \((-\frac{5}{2},-\frac{5}{4})\). 3. Differentiate: \(f'(x)=2x+3\) and \(g'(x)=-2x\). Equal slopes require \(2x+3=-2x\), so \(x=-\frac{3}{4}\). 4. At \(x=2\), \(g(2)=1\) and the tangent slope is \(-4\). The normal slope is \(\frac{1}{4}\). 5. The normal line is \(y-1=\frac{1}{4}(x-2)\), or \(y=\frac{1}{4}x+\frac{1}{2}\).

a) \((1,4)\) and \((-\frac{5}{2},-\frac{5}{4})\) b) \(x=-\frac{3}{4}\) c) \(y=\frac{1}{4}x+\frac{1}{2}\)

- A tangent parallel to the x-axis has slope \(0\). - Write the tangent line at \(x=a\) using point-slope form. - Treat \(a\) as a fixed parameter while simplifying. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(h'(x)=\frac{1}{2}x-1\). A horizontal tangent has slope \(0\), so \(\frac{1}{2}x-1=0\), giving \(x=2\). 2. Since \(h(2)=-1\), the point is \(P=(2,-1)\). 3. At \(x=a\), the slope is \(\frac{1}{2}a-1\), and the point is \((a,\frac{1}{4}a^2-a)\). 4. Point-slope form gives \(y=(\frac{1}{2}a-1)(x-a)+\frac{1}{4}a^2-a\). 5. Simplifying gives \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\). 6. Setting \(x=0\) gives \(y=-\frac{1}{4}a^2\), so the y-intercept is \((0,-\frac{1}{4}a^2)\).

a) \(P=(2,-1)\) b) \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\) c) Setting \(x=0\) gives \(S=(0,-\frac{1}{4}a^2)\).

- Set the derivative equal to the required slope. - Solve for every possible input. - Substitute each input into the original function. - Use the common slope and each point to write the tangent lines.

1. Differentiate: \(f'(x)=3x^2-6x-6\). 2. Set the derivative equal to \(3\): \(3x^2-6x-6=3\). Dividing by \(3\) gives \(x^2-2x-3=0\). 3. Factor: \((x-3)(x+1)=0\), so \(x=3\) or \(x=-1\). 4. Evaluate the function: \(f(3)=-16\) and \(f(-1)=4\). The points are \((3,-16)\) and \((-1,4)\). 5. At \((3,-16)\), the tangent is \(y+16=3(x-3)\), or \(y=3x-25\). 6. At \((-1,4)\), the tangent is \(y-4=3(x+1)\), or \(y=3x+7\).

The points are \((3,-16)\) and \((-1,4)\). The tangent lines are \(y=3x-25\) and \(y=3x+7\), respectively.

- What must a function value equal where its graph crosses the x-axis? - Recall the power rule, including its use with negative exponents. - What relationship between two nonzero slopes indicates perpendicular tangent lines? - Compare the products of the derivatives at the common point.

1. Solve \(0.5x^2-2=0\): \(x^2=4\), so \(x=\pm2\). The positive zero is \(x_0=2\). 2. Evaluate the other functions at \(x=2\): \(g(2)=-0.125\cdot2^2+0.5=0\) and \(h(2)=\frac{2}{2}-1=0\). Therefore, all three graphs pass through \((2,0)\). 3. Differentiate: \(f'(x)=x\), \(g'(x)=-0.25x\), and \(h'(x)=-\frac{2}{x^2}\). At \(x=2\), the slopes are \(f'(2)=2\) and \(g'(2)=h'(2)=-0.5\). Since \(2\cdot(-0.5)=-1\), each of the other two graphs intersects the graph of \(f\) at a right angle.

1. \(x_0=2\) 2. Yes. All three graphs cross the x-axis at \((2,0)\). 3. Yes. The products of the tangent slopes are \(f'(2)g'(2)=-1\) and \(f'(2)h'(2)=-1\).

- A common tangent requires both a common point and a common derivative value. - Look for an input that makes the term containing \(k\) equal to zero. - Use the point and slope to write the tangent line.

1. For the function value to be independent of \(k\), the parameter term must be zero. Since \((x-2)^3=0\), use \(x=2\). 2. Evaluate the function: \(g_k(2)=k(2-2)^3+2^2+1=5\). Every graph passes through \((2,5)\). 3. Differentiate: \(g_k'(x)=3k(x-2)^2+2x\). 4. Evaluate the derivative at \(x=2\): \(g_k'(2)=4\), independent of \(k\). 5. The line with slope \(4\) through \((2,5)\) is \(y-5=4(x-2)\), or \(y=4x-3\).

Every graph has the same tangent line at \((2,5)\). Its equation is \(y=4x-3\).

- Tangent graphs have the same function value and derivative at the point of tangency. - Start with the general form of a quadratic polynomial. - Use the point \(A\) to determine the remaining coefficient. - Check that the leading coefficient is nonzero.

1. Write \(p(x)=ax^2+bx+c\). 2. Since the graphs are tangent at the origin, they have the same value and slope there. Because \(f(0)=0\), \(p(0)=0\), so \(c=0\). 3. Since \(f'(x)=\frac{1}{x+1}\), \(f'(0)=1\). Also, \(p'(x)=2ax+b\), so \(p'(0)=b=1\). 4. Thus \(p(x)=ax^2+x\). Using \(p(2)=4\), \(4a+2=4\), so \(a=\frac{1}{2}\). 5. Therefore, \(p(x)=\frac{1}{2}x^2+x\).

\(p(x)=\frac{1}{2}x^2+x\)

- Start with the general form of a cubic polynomial. - Tangency at \(x=0\) gives one condition for the value and one for the slope. - A horizontal tangent means the derivative is zero. - Use all conditions to form and solve a linear system.

1. Write \(p(x)=ax^3+bx^2+cx+d\), so \(p'(x)=3ax^2+2bx+c\). 2. Tangency at \(x=0\) requires equal function values and slopes. Since \(f(0)=4\), \(d=4\). 3. Differentiate \(f\): \(f'(x)=-2e^{-0.5x}\), so \(f'(0)=-2\). Therefore, \(c=-2\). 4. The horizontal tangent at \(x=-1\) gives \(p'(-1)=3a-2b-2=0\), so \(3a-2b=2\). 5. Since \(p(1)=2\), \(a+b-2+4=2\), so \(a+b=0\). 6. Solving the system gives \(a=\frac{2}{5}\) and \(b=-\frac{2}{5}\). 7. Therefore, \(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\).

\(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\)

- A tangent line through the origin has equation \(y=mx\). - At the point of tangency, use \(m=f'(x_0)\) and \(f(x_0)=mx_0\). - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Differentiate: \(f'(x)=\frac{3}{2}x^2-4x+1\). 2. A tangent line through the origin at \(x=x_0\) must satisfy \(f(x_0)=f'(x_0)x_0\). 3. Substitute: \(\frac{1}{2}x_0^3-2x_0^2+x_0=x_0\left(\frac{3}{2}x_0^2-4x_0+1\right)\). 4. Simplifying gives \(x_0^2(x_0-2)=0\). The solutions are \(x_0=0\) and \(x_0=2\). Since the point of tangency is not the origin, \(x_0=2\). 5. Since \(f(2)=-2\), the point of tangency is \(B(2, -2)\). The tangent slope is \(f'(2)=-1\), so the tangent line is \(t: y=-x\). 6. At the origin, the tangent slope is \(f'(0)=1\). Because \((-1)(1)=-1\), the two tangent lines are perpendicular.

a) \(B(2, -2)\); \(t: y=-x\) b) The slopes are \(-1\) and \(1\), whose product is \(-1\), so the tangent lines are perpendicular.

- Set the two function expressions equal to find their intersection points. - The derivative gives the tangent slope at an input. - Relate the slope of a line to its angle of inclination. - Use the acute-angle formula for two lines, and make sure the result is at most \(90^\circ\).

1. Set the functions equal: \(x^2-6x+9=-x+5\). Then \(x^2-5x+4=0\), so \((x-1)(x-4)=0\). Thus, \(x=1\) or \(x=4\). 2. Using \(g(x)=-x+5\), the intersection points are \(P_1(1,4)\) and \(P_2(4,1)\). 3. Differentiate: \(f'(x)=2x-6\) and \(g'(x)=-1\). At \(P_1\), the slopes are \(f'(1)=-4\) and \(g'(1)=-1\). At \(P_2\), the slopes are \(f'(4)=2\) and \(g'(4)=-1\). 4. For two lines with slopes \(m_1\) and \(m_2\), \(\tan(\alpha)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\). At \(P_1\), \(\tan(\alpha_1)=\left|\frac{-4-(-1)}{1+(-4)(-1)}\right|=0.6\), so \(\alpha_1\approx30.96^\circ\). 5. At \(P_2\), \(\tan(\alpha_2)=\left|\frac{2-(-1)}{1+2(-1)}\right|=3\), so \(\alpha_2\approx71.57^\circ\).

a) \(P_1(1,4)\) and \(P_2(4,1)\) b) At \(P_1\), the slopes are \(-4\) for \(f\) and \(-1\) for \(g\). At \(P_2\), the slopes are \(2\) for \(f\) and \(-1\) for \(g\). c) \(\alpha_1\approx30.96^\circ\) and \(\alpha_2\approx71.57^\circ\)

- Write the tangent-line equation at a general input. - A line through \(Q(0,7)\) must have a y-value of \(7\) when \(x=0\). - Express the tangent slope using the derivative and also using the point of tangency and \(Q\). - Equate the two slope expressions to solve for the possible inputs.

1. The tangent line at \(x=u\) is \(t(x)=f'(u)(x-u)+f(u)\). 2. Differentiate: \(f'(x)=-0.5x+1\). Therefore, \(t(x)=(-0.5u+1)(x-u)+(-0.25u^2+u+3)\). 3. Since the tangent passes through \(Q(0,7)\), set \(t(0)=7\): \((-0.5u+1)(-u)-0.25u^2+u+3=7\). This simplifies to \(0.25u^2=4\), so \(u=4\) or \(u=-4\). 4. Evaluate the function: \(f(4)=3\) and \(f(-4)=-5\). The points of tangency are \(B_1(4,3)\) and \(B_2(-4,-5)\). 5. The slopes are \(f'(4)=-1\) and \(f'(-4)=3\). Thus, \(t_1\) is \(y-3=-(x-4)\), or \(y=-x+7\), and \(t_2\) is \(y+5=3(x+4)\), or \(y=3x+7\).

a) \(B_1(4,3)\) and \(B_2(-4,-5)\) b) \(t_1: y=-x+7\) and \(t_2: y=3x+7\)

- The slope of the graph at an input is the derivative value there. - Apply the power rule to find the derivative. - For part b), set the derivative equal to the given slope. - Parallel lines have equal slopes. - After finding each x-coordinate, evaluate the original function to obtain the full point.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-2x\). 2. Set the derivative equal to \(-2\): \(\frac{1}{2}a^2-2a=-2\). This gives \(a^2-4a+4=0\), so \(a=2\). 3. Evaluate the function: \(f(2)=\frac{1}{6}\cdot2^3-2^2+2=-\frac{2}{3}\). Thus, \(A\left(2,-\frac{2}{3}\right)\). 4. A line parallel to \(y=6x-1\) has slope \(6\). Set \(\frac{1}{2}x^2-2x=6\). Then \(x^2-4x-12=0\), so \((x-6)(x+2)=0\). Therefore, \(x=6\) or \(x=-2\). 5. Evaluate the function: \(f(6)=2\) and \(f(-2)=-\frac{10}{3}\). The points are \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\).

a) \(f'(x)=\frac{1}{2}x^2-2x\) b) \(A\left(2,-\frac{2}{3}\right)\) c) \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\)

- Read two exact points from the graph and substitute them into \(f(x)=ax^2+bx\). - Evaluate \(f(3)\) and use the derivative for the tangent slope. - Convert the tangent slope to the normal slope. - Use the normal line's axis intercepts as the triangle's legs.

1. From the graph, use \((2,2)\) and \((4,0)\). Substitution gives \(4a+2b=2\) and \(16a+4b=0\). 2. Simplifying gives \(2a+b=1\) and \(4a+b=0\). Subtracting yields \(a=-0.5\), and then \(b=2\). Thus, \(f(x)=-0.5x^2+2x\). 3. Evaluate the function: \(f(3)=1.5\), so \(P=(3,1.5)\). Differentiate: \(f'(x)=-x+2\), so the tangent slope at \(x=3\) is \(-1\). 4. The normal slope is \(1\). Through \(P\), the normal is \(y-1.5=x-3\), or \(y=x-1.5\). 5. Its intercepts are \((0,-1.5)\) and \((1.5,0)\). The triangle area is \(\frac{1}{2}(1.5)(1.5)=1.125=\frac{9}{8}\) square units.

a) \(a=-0.5\), \(b=2\) b) \(n: y=x-1.5\) c) \(\frac{9}{8}\) square units

- First compute \(f'(x)\) and \(g'(x)\). - Compare each graph's degree, symmetry, vertex, and intercepts. - Check that the zeros of each derivative match horizontal tangents of its function.

1. Differentiate: \(f'(x) = x\) and \(g'(x) = -0.6x^2 + 1.5\). 2. Graph (3) is an upward-opening parabola with vertex \((0, -2)\), so it represents \(f\). Graph (1) is the line \(y = x\), so it represents \(f'\). 3. Graph (2) is an odd cubic with negative leading coefficient, so it represents \(g\). Graph (4) is a downward-opening parabola with vertex \((0, 1.5)\), so it represents \(g'\).

(1) \(f'\) (2) \(g\) (3) \(f\) (4) \(g'\)

- Set the function equal to zero to find the x-intercepts. - Use the derivative to find each tangent slope. - Use a point and its slope to write each tangent line. - Solve the two tangent equations as a system to find \(S\). - Use the triangle area formula with \(\overline{PQ}\) as the base.

1. Set \(f(x)=0\): \(-0.25x^2+4=0\), so \(x^2=16\) and \(x=\pm4\). Thus, \(P=(-4,0)\) and \(Q=(4,0)\). 2. Differentiate: \(f'(x)=-0.5x\). At \(P\), the slope is \(f'(-4)=2\), so \(t_1\) is \(y=2(x+4)=2x+8\). 3. At \(Q\), the slope is \(f'(4)=-2\), so \(t_2\) is \(y=-2(x-4)=-2x+8\). 4. Set the tangent equations equal: \(2x+8=-2x+8\), giving \(x=0\) and \(y=8\). Thus, \(S=(0,8)\). 5. Triangle \(PQS\) has base \(8\) and height \(8\). Its area is \(\frac{1}{2}\cdot8\cdot8=32\) square units.

a) \(P=(-4,0)\) and \(Q=(4,0)\) b) \(t_1: y=2x+8\) and \(t_2: y=-2x+8\) c) \(32\) square units

- Identify the vertex and zeros of \(f\). - Adding outside the function shifts a graph vertically; adding inside the input shifts it horizontally. - The derivative of a quadratic function is linear.

1. The graph of \(f(x) = x^2 - 2x\) is an upward-opening parabola with zeros at \(x = 0\) and \(x = 2\) and vertex \((1, -1)\). This is graph II. 2. The derivative is \(f'(x) = 2x - 2\), a line with slope \(2\) and \(y\)-intercept \(-2\). This is graph IV. 3. The graph of \(g(x) = f(x) + 3\) is shifted up \(3\) units, so its vertex is \((1, 2)\). This is graph I. 4. The graph of \(h(x) = f(x + 2)\) is shifted left \(2\) units, so its vertex is \((-1, -1)\). This is graph III.

Graph I: \(g(x) = f(x) + 3\) Graph II: \(f(x)\) Graph III: \(h(x) = f(x + 2)\) Graph IV: \(f'(x)\)

- Set the function expressions equal to find the intersections. - The angle between the curves is the angle between their tangent lines. - Use derivatives to find the tangent slopes. - Apply the formula for the acute angle between two lines.

1. Set the functions equal: \(x^2-1=7-x^2\). Then \(2x^2=8\), so \(x=\pm2\). 2. Since \(f(2)=f(-2)=3\), the intersection points are \((2,3)\) and \((-2,3)\). 3. Differentiate: \(f'(x)=2x\) and \(g'(x)=-2x\). At \(x=2\), the tangent slopes are \(4\) and \(-4\). 4. Use \(\tan(\phi)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\): \(\tan(\phi)=\left|\frac{4-(-4)}{1+4(-4)}\right|=\frac{8}{15}\). Thus, \(\phi=\arctan\left(\frac{8}{15}\right)\approx28.1^\circ\). 5. By symmetry, the acute angle at the other intersection point is also approximately \(28.1^\circ\).

a) \((2,3)\) and \((-2,3)\) b) The angle at each intersection is approximately \(28.1^\circ\).

- Use the derivative at \(x=u\) to find the tangent slope. - Write the tangent equation in point-slope form. - Set \(x=0\) to find the y-intercept. - The projection \(P_1\) has x-coordinate \(0\) and the same y-coordinate as \(P\).

1. Differentiate: \(f'(x)=2x\). 2. At \(x=u\), the tangent line is \(y=2u(x-u)+u^2=2ux-u^2\). 3. At the y-axis, \(x=0\), so \(y=-u^2\). Therefore, the y-intercept is \(Q(0,-u^2)\). 4. The projection point is \(P_1(0,u^2)\). 5. The distance from \(Q\) to the origin is \(|-u^2|=u^2\), and the distance from \(P_1\) to the origin is \(|u^2|=u^2\). 6. The distances are equal, so the origin is the midpoint of \(\overline{P_1Q}\).

a) \(t: y=2ux-u^2\), so its y-intercept is \(Q(0,-u^2)\). b) Both distances equal \(u^2\). The origin is the midpoint of \(\overline{P_1Q}\).

- Differentiate \(f\) and \(g\) first. - Compare the degree and symmetry of each derivative with the graphs. - Find the zeros of each derivative and match them to the \(x\)-intercepts shown.

1. Differentiate each function using the power rule: \(f'(x) = x^2 - 1\) and \(g'(x) = x^3 - 4x\). 2. The graph of \(f'\) is an upward-opening parabola, symmetric about the \(y\)-axis, with vertex \((0, -1)\) and zeros at \(x=-1\) and \(x=1\). This is graph A. 3. Since \(g'(x)=x(x-2)(x+2)\), its graph is an odd cubic with zeros at \(x=-2\), \(x=0\), and \(x=2\). This is graph B.

Graph A represents \(f'\), and graph B represents \(g'\).

- Use the slope formula for the two known points. - Write the tangent equation at a general input \(a\). - A point lies on a line when its coordinates satisfy the line equation. - Use the derivative to express the tangent slope.

1. The point is \(Q_a\left(a,\frac{1}{2}a^2+2\right)\). 2. The secant slope is \(m_a=\frac{f(a)-0}{a-0}=\frac{\frac{1}{2}a^2+2}{a}=\frac{a^2+4}{2a}\). 3. Differentiate: \(f'(x)=x\). 4. The tangent at \(Q_a\) is \(y=a(x-a)+\frac{1}{2}a^2+2=ax-\frac{1}{2}a^2+2\). 5. For this line to pass through \(P=(0,0)\), \(0=-\frac{1}{2}a^2+2\). Thus, \(a^2=4\), so \(a=2\) or \(a=-2\).

a) \(m_a=\frac{a^2+4}{2a}\) b) \(a=2\) and \(a=-2\)

- Rewrite the function using a negative exponent before differentiating. - Use point-slope form at \(x=a\). - A point on the y-axis has x-coordinate \(0\). - Substitute the coordinates of \(P\) into the tangent-line equation.

1. Differentiate: \(g'(x)=-\frac{2}{x^2}\). 2. The tangent line at \(x=a\) is \(y=g'(a)(x-a)+g(a)=-\frac{2}{a^2}(x-a)+\frac{2}{a}\). 3. Simplify: \(t_a: y=-\frac{2}{a^2}x+\frac{4}{a}\). 4. Since the line passes through \(P(0, 4)\), \(4=\frac{4}{a}\). Therefore, \(a=1\).

a) \(t_a: y=-\frac{2}{a^2}x+\frac{4}{a}\) b) \(a=1\)

- Use the factored form of a quadratic with the given zeros. - Differentiate to find the tangent slope at the origin. - Write the tangent line through the origin. - Substitute the tangent lines' intersection point to determine the remaining coefficient.

1. Since the zeros are \(0\) and \(4\), write \(f(x)=ax(x-4)=ax^2-4ax\). 2. Differentiate: \(f'(x)=2ax-4a\). At the origin, the tangent slope is \(f'(0)=-4a\), so the tangent line there is \(y=-4ax\). 3. The point \(S(2, -6)\) lies on this tangent line. Therefore, \(-6=-4a(2)=-8a\), so \(a=\frac{3}{4}\). 4. Thus \(f(x)=\frac{3}{4}x(x-4)=\frac{3}{4}x^2-3x\).

\(f(x)=\frac{3}{4}x^2-3x\)

- Write a general tangent line at an unknown input \(u\). - The tangent slope at \(u\) is \(f'(u)\). - Require the tangent line to pass through the given external point. - Use the condition \(x>0\) when choosing a solution.

1. Differentiate: \(f'(x)=\frac{1}{4}x\). 2. Let the point of tangency have x-coordinate \(u\). The tangent line is \(y=f'(u)(x-u)+f(u)\). 3. Substitute the expressions for \(f'(u)\) and \(f(u)\): \(y=\frac{1}{4}u(x-u)+\frac{1}{8}u^2+2=\frac{1}{4}ux-\frac{1}{8}u^2+2\). 4. The tangent passes through \((0,0)\), so \(0=-\frac{1}{8}u^2+2\). Thus, \(u^2=16\). Because \(u>0\), \(u=4\). 5. Find the point and slope: \(f(4)=4\) and \(f'(4)=1\). Therefore, \(B=(4,4)\) and the ray has slope \(1\).

\(B=(4,4)\) and \(m=1\)

- A tangent at input \(u\) has slope \(h'(u)\). - Write the tangent line at the unknown point of tangency. - Require that line to pass through \(P=(2,0)\). - The resulting equation may have more than one solution for \(u\).

1. Differentiate: \(h'(x)=2x-4\). 2. Let the point of tangency be \(B=(u,h(u))\). A tangent line at \(u\) is \(y=(2u-4)(x-u)+u^2-4u+5\). 3. Require the line to pass through \(P=(2,0)\): \(0=(2u-4)(2-u)+u^2-4u+5\). 4. Simplifying gives \(u^2-4u+3=0\), so \(u=1\) or \(u=3\). 5. The points of tangency are \((1,h(1))=(1,2)\) and \((3,h(3))=(3,2)\). 6. At \(u=1\), the slope is \(-2\), giving \(y=-2(x-2)=-2x+4\). At \(u=3\), the slope is \(2\), giving \(y=2(x-2)=2x-4\).

The points of tangency are \((1,2)\) and \((3,2)\). The tangent lines are \(y=-2x+4\) and \(y=2x-4\), respectively.

- First find the two x-intercepts of the parabola. - Use the derivative to write both tangent lines. - Their intersection is the third vertex of the triangle. - Use the x-axis segment as the base when computing the area.

1. Factor \(f_k(x)=x(x-k)\). The x-intercepts are \(O=(0,0)\) and \(A=(k,0)\). 2. Differentiate: \(f_k'(x)=2x-k\). The tangent slopes are \(f_k'(0)=-k\) and \(f_k'(k)=k\). 3. The tangent lines are \(y=-kx\) and \(y=k(x-k)=kx-k^2\). 4. Set the equations equal: \(-kx=kx-k^2\). Since \(k>0\), \(x=\frac{k}{2}\). Then \(y=-\frac{k^2}{2}\), so \(S=(\frac{k}{2},-\frac{k^2}{2})\). 5. The base \(OA\) has length \(k\), and the height is \(\frac{k^2}{2}\). Thus, the area is \(\frac{1}{2}\cdot k\cdot\frac{k^2}{2}=\frac{k^3}{4}\). 6. Solve \(\frac{k^3}{4}=16\). This gives \(k^3=64\), so \(k=4\).

\(k=4\)

- Use the slope formula for two given points. - The derivative at \(a\) gives the tangent slope at \(Q_a\). - Check \(a=0\) separately before using a slope formula that divides by \(a\). - If the tangent passes through \(P\), compare its slope with the slope of \(PQ_a\). - Solve the resulting equation for \(a\).

1. The slope through \(P=(0,-5)\) and \(Q_a=(a,a^2-1)\) is \(m_a=\frac{(a^2-1)-(-5)}{a-0}=\frac{a^2+4}{a}\). 2. Differentiate: \(f'(x)=2x\), so the tangent slope at \(Q_a\) is \(f'(a)=2a\). 3. If \(a=0\), the tangent line is \(y=-1\), which does not pass through \(P\). For \(a\ne 0\), a tangent line through both \(Q_a\) and \(P\) must have slope \(m_a\). Thus, \(2a=\frac{a^2+4}{a}\). 4. Since \(a\ne 0\), multiply by \(a\): \(2a^2=a^2+4\). Therefore, \(a^2=4\), so \(a=-2\) or \(a=2\).

a) \(m_a=\frac{a^2+4}{a}\) b) \(a=-2\) and \(a=2\)

- Use the slope formula for two points. - The derivative at \(a\) gives the tangent slope. - Check \(a=0\) separately before using a slope formula that divides by \(a\). - If the tangent passes through \(R\), compare its slope with the slope of \(RS_a\). - Solve the resulting equation for \(a\).

1. Use the slope formula: \(m_a=\frac{(\frac{1}{2}a^3+4)-(-4)}{a-0}=\frac{\frac{1}{2}a^3+8}{a}=\frac{a^3+16}{2a}\). 2. Differentiate: \(g'(x)=\frac{3}{2}x^2\), so the tangent slope at \(x=a\) is \(\frac{3}{2}a^2\). 3. If \(a=0\), the tangent line is \(y=4\), which does not pass through \(R\). For \(a\ne 0\), a tangent line through both \(S_a\) and \(R\) must have slope \(m_a\): \(\frac{3}{2}a^2=\frac{a^3+16}{2a}\). 4. Since \(a\ne 0\), multiply by \(2a\): \(3a^3=a^3+16\). Thus, \(2a^3=16\), so \(a=2\).

a) \(m_a=\frac{a^3+16}{2a}\) b) \(a=2\)

- Treat \(k\) as a constant when differentiating and simplifying. - Use the point-slope form of a tangent line at \(x=x_0\). - For the triangle, find the length of the side on the x-axis and the vertical distance from the third vertex to the x-axis. - Lengths and areas must be positive.

1. Solve \(x^3 - kx = 0\): \(x(x^2-k)=0\), so the zeros are \(x=0\) and \(x=\pm\sqrt{k}\). Therefore, \(x_P=\sqrt{k}\). 2. Differentiate: \(f_k'(x)=3x^2-k\). 3. At the origin, \(f_k'(0)=-k\). Since the tangent passes through \((0, 0)\), \(t_0: y=-kx\). 4. At \(P(\sqrt{k}, 0)\), the slope is \(f_k'(\sqrt{k})=3k-k=2k\). Thus \(t_P: y=2k(x-\sqrt{k})\). 5. Find the intersection of the tangent lines by solving \(-kx=2k(x-\sqrt{k})\). Since \(k>0\), \(x=\frac{2}{3}\sqrt{k}\), and then \(y=-\frac{2}{3}k\sqrt{k}\). 6. The triangle has base \(\sqrt{k}\) and height \(\frac{2}{3}k\sqrt{k}\). Therefore, \(A(k)=\frac{1}{2}\cdot\sqrt{k}\cdot\frac{2}{3}k\sqrt{k}=\frac{1}{3}k^2\).

a) \(x_P=\sqrt{k}\) b) \(t_0: y=-kx\); \(t_P: y=2k(x-\sqrt{k})\) c) The tangent lines intersect at \(\left(\frac{2}{3}\sqrt{k}, -\frac{2}{3}k\sqrt{k}\right)\), giving \(A(k)=\frac{1}{3}k^2\).

- Use the normal-line formula at a general input \(a\). - Differentiate the square-root function. - Set \(y=0\) to find the x-intercept. - A constant difference does not depend on \(a\).

1. Differentiate: \(f'(x)=\frac{1}{2\sqrt{x}}\). 2. At \(x=a\), the normal slope is \(m_n=-\frac{1}{f'(a)}=-2\sqrt{a}\). 3. The normal line is \(y-\sqrt{a}=-2\sqrt{a}(x-a)\). 4. At the x-intercept, \(y=0\): \(0=-2\sqrt{a}(x-a)+\sqrt{a}\). Since \(a>0\), divide by \(\sqrt{a}\) to get \(2(x-a)=1\). 5. Thus, \(x=a+0.5\), and the horizontal distance is \((a+0.5)-a=0.5\).

a) \(n_a: y=-2\sqrt{a}(x-a)+\sqrt{a}\) b) \(S_x=(a+0.5,0)\), so the horizontal distance is always \(0.5\).

- Write the tangent line at an unknown input \(u\). - Require that line to pass through the external point. - Use the midpoint formula after finding both points of tangency. - Points on the same vertical line have distance equal to the difference of their y-coordinates.

1. A tangent at input \(u\) has equation \(y=f'(u)(x-u)+f(u)\). Since \(f'(x)=2x\), this becomes \(y=2u(x-u)+u^2-2=2ux-u^2-2\). 2. Require the tangent to pass through \(P=(0,-3)\): \(-3=-u^2-2\). Thus, \(u^2=1\), so \(u=-1\) or \(u=1\). 3. The points of tangency are \((-1,-1)\) and \((1,-1)\). 4. For \(u=-1\), the tangent is \(y=-2x-3\). For \(u=1\), the tangent is \(y=2x-3\). 5. The midpoint is \(M=(0,-1)\). Since \(M\) and \(P\) have the same x-coordinate, their distance is \(|-1-(-3)|=2\).

a) \(B_1=(-1,-1)\) and \(B_2=(1,-1)\) b) \(y=-2x-3\) and \(y=2x-3\) c) \(M=(0,-1)\), and the distance from \(M\) to \(P\) is \(2\).

- Relate the tangent and normal slopes at a general point. - Write the normal line at \(x=u\). - Substitute the coordinates of \(A\). - Check separately the case where the tangent is horizontal.

1. Since \(P\) lies on the graph, \(v=u^2\). Also, \(f'(x)=2x\). 2. For \(u\ne0\), the normal slope is \(-\frac{1}{2u}\), so the normal line is \(y-u^2=-\frac{1}{2u}(x-u)\). 3. Substitute \(A(0,1.5)\): \(1.5-u^2=\frac{1}{2}\). Thus, \(u^2=1\), so \(u=1\) or \(u=-1\). These give \((1,1)\) and \((-1,1)\). 4. Check \(u=0\) separately. The tangent at \((0,0)\) is horizontal, so the normal is the y-axis, which contains \(A\). Therefore, \((0,0)\) is also a solution.

\((1,1)\), \((-1,1)\), and \((0,0)\)

- Apply the power rule to \(x^{-1}\). - Convert the tangent slope to the normal slope. - Write the normal line at a general input and substitute the origin. - Find all real solutions of \(a^4=1\).

1. Differentiate: \(f'(x)=-\frac{1}{x^2}\). 2. At \(x=a\), the normal slope is \(m_n=-\frac{1}{f'(a)}=a^2\). 3. The normal line through \(P\left(a,\frac{1}{a}\right)\) is \(y-\frac{1}{a}=a^2(x-a)\). 4. Substitute the origin: \(-\frac{1}{a}=-a^3\), so \(a^4=1\). 5. The real solutions are \(a=1\) and \(a=-1\). The corresponding points are \((1,1)\) and \((-1,-1)\).

\(a=1\) gives \(P=(1,1)\), and \(a=-1\) gives \(P=(-1,-1)\).

- Use the tangent-line formula at a general input \(a\). - The derivative gives the tangent slope. - Substitute the coordinates of \(P\) into the general tangent equation. - Factor the resulting equation in \(a\).

1. Differentiate: \(f'(x)=1.5x^2-1.5\). 2. The tangent at \(x=a\) is \(t_a(x)=f'(a)(x-a)+f(a)\). Substitution gives \(t_a(x)=(1.5a^2-1.5)(x-a)+0.5a^3-1.5a\), which simplifies to \(t_a(x)=(1.5a^2-1.5)x-a^3\). 3. Require the tangent to pass through \(P(2,-3)\): \(-3=2(1.5a^2-1.5)-a^3\). This simplifies to \(a^2(3-a)=0\). 4. Therefore, \(a=0\) or \(a=3\). 5. Substitution gives \(t_0(x)=-1.5x\) and \(t_3(x)=12x-27\).

a) \(t_a(x)=(1.5a^2-1.5)x-a^3\) b) \(a=0\) gives \(t_0(x)=-1.5x\), and \(a=3\) gives \(t_3(x)=12x-27\).

- Apply the power rule after writing the reciprocal with a negative exponent. - Find the tangent line's x- and y-intercepts. - The intercepts give the legs of the right triangle. - Check whether \(a\) remains after simplifying the area formula.

1. Differentiate: \(f'(x)=-\frac{1}{x^2}\). 2. At \(x=a\), the tangent line is \(y=-\frac{1}{a^2}(x-a)+\frac{1}{a}=-\frac{1}{a^2}x+\frac{2}{a}\). 3. Setting \(x=0\) gives the y-intercept \(S_y\left(0,\frac{2}{a}\right)\). 4. Setting \(y=0\) gives \(x=2a\), so the x-intercept is \(S_x(2a,0)\). 5. The triangle has base \(2a\) and height \(\frac{2}{a}\). Its area is \(\frac{1}{2}(2a)\left(\frac{2}{a}\right)=2\), so it is independent of \(a\).

a) \(y=-\frac{1}{a^2}x+\frac{2}{a}\) b) The intercepts are \(S_x(2a,0)\) and \(S_y\left(0,\frac{2}{a}\right)\), and the area is always \(2\).