Calculus Examples

$f (x) = - \frac{5}{3} x^{2} + 2 x + 2$ ; $(- 1, - \frac{5}{3})$

Step 1

Find the first derivative and evaluate at $x = - 1$ and $y = - \frac{5}{3}$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of $- \frac{5}{3} x^{2} + 2 x + 2$ with respect to $x$ is $\frac{d}{d x} [- \frac{5}{3} x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [- \frac{5}{3} x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2

Evaluate $\frac{d}{d x} [- \frac{5}{3} x^{2}]$ .

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Step 1.2.1

Since $- \frac{5}{3}$ is constant with respect to $x$ , the derivative of $- \frac{5}{3} x^{2}$ with respect to $x$ is $- \frac{5}{3} \frac{d}{d x} [x^{2}]$ .

$- \frac{5}{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- \frac{5}{3} (2 x) + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.3

Multiply $2$ by $- 1$ .

$- 2 (\frac{5}{3}) x + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.4

Combine $- 2$ and $\frac{5}{3}$ .

$\frac{- 2 \cdot 5}{3} x + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.5

Multiply $- 2$ by $5$ .

$\frac{- 10}{3} x + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.6

Combine $\frac{- 10}{3}$ and $x$ .

$\frac{- 10 x}{3} + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.2.7

Move the negative in front of the fraction.

$- \frac{10 x}{3} + \frac{d}{d x} [2 x] + \frac{d}{d x} [2]$

Step 1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$- \frac{10 x}{3} + 2 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- \frac{10 x}{3} + 2 \cdot 1 + \frac{d}{d x} [2]$

Step 1.3.3

Multiply $2$ by $1$ .

$- \frac{10 x}{3} + 2 + \frac{d}{d x} [2]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$- \frac{10 x}{3} + 2 + 0$

Step 1.4.2

Add $- \frac{10 x}{3} + 2$ and $0$ .

$- \frac{10 x}{3} + 2$

Step 1.5

Evaluate the derivative at $x = - 1$ .

$- \frac{10 (- 1)}{3} + 2$

Step 1.6

Simplify.

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Step 1.6.1

Simplify each term.

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Step 1.6.1.1

Multiply $10$ by $- 1$ .

$- \frac{- 10}{3} + 2$

Step 1.6.1.2

Move the negative in front of the fraction.

$- - \frac{10}{3} + 2$

Step 1.6.1.3

Multiply $- - \frac{10}{3}$ .

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Step 1.6.1.3.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{10}{3}) + 2$

Step 1.6.1.3.2

Multiply $\frac{10}{3}$ by $1$ .

$\frac{10}{3} + 2$

Step 1.6.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{10}{3} + 2 \cdot \frac{3}{3}$

Step 1.6.3

Combine $2$ and $\frac{3}{3}$ .

$\frac{10}{3} + \frac{2 \cdot 3}{3}$

Step 1.6.4

Combine the numerators over the common denominator.

$\frac{10 + 2 \cdot 3}{3}$

Step 1.6.5

Simplify the numerator.

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Step 1.6.5.1

Multiply $2$ by $3$ .

$\frac{10 + 6}{3}$

Step 1.6.5.2

Add $10$ and $6$ .

$\frac{16}{3}$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $\frac{16}{3}$ and a given point $(- 1, - \frac{5}{3})$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- \frac{5}{3}) = \frac{16}{3} \cdot (x - (- 1))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y + \frac{5}{3} = \frac{16}{3} \cdot (x + 1)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $\frac{16}{3} \cdot (x + 1)$ .

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Step 2.3.1.1

Rewrite.

$y + \frac{5}{3} = 0 + 0 + \frac{16}{3} \cdot (x + 1)$

Step 2.3.1.2

Simplify by adding zeros.

$y + \frac{5}{3} = \frac{16}{3} \cdot (x + 1)$

Step 2.3.1.3

Apply the distributive property.

$y + \frac{5}{3} = \frac{16}{3} x + \frac{16}{3} \cdot 1$

Step 2.3.1.4

Combine $\frac{16}{3}$ and $x$ .

$y + \frac{5}{3} = \frac{16 x}{3} + \frac{16}{3} \cdot 1$

Step 2.3.1.5

Multiply $\frac{16}{3}$ by $1$ .

$y + \frac{5}{3} = \frac{16 x}{3} + \frac{16}{3}$

Step 2.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 2.3.2.1

Subtract $\frac{5}{3}$ from both sides of the equation.

$y = \frac{16 x}{3} + \frac{16}{3} - \frac{5}{3}$

Step 2.3.2.2

Combine the numerators over the common denominator.

$y = \frac{16 x + 16 - 5}{3}$

Step 2.3.2.3

Subtract $5$ from $16$ .

$y = \frac{16 x + 11}{3}$

Step 2.3.2.4

Split the fraction $\frac{16 x + 11}{3}$ into two fractions.

$y = \frac{16 x}{3} + \frac{11}{3}$

Step 2.3.3

Reorder terms.

$y = \frac{16}{3} x + \frac{11}{3}$

Step 3