Calculus Examples

$y = x^{2} + 4 x - 8$

Set $y$ as a function of $x$ .

$f (x) = x^{2} + 4 x - 8$

Find the derivative.

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By the Sum Rule, the derivative of $x^{2} + 4 x - 8$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 8]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 8]$

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

$2 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 8]$

Evaluate $\frac{d}{d x} [4 x]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$2 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [- 8]$

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

$2 x + 4 \cdot 1 + \frac{d}{d x} [- 8]$

Multiply $4$ by $1$ to get $4$ .

$2 x + 4 + \frac{d}{d x} [- 8]$

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$2 x + 4 + 0$

Add $2 x + 4$ and $0$ to get $2 x + 4$ .

$2 x + 4$

Set the derivative equal to $0$ then solve the equation $2 x + 4 = 0$ .

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Subtract $4$ from both sides of the equation.

$2 x = - 4$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = - 4$ by $2$ .

$\frac{2 x}{2} = - \frac{4}{2}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{2 x}{2} = - \frac{4}{2}$

Divide $x$ by $1$ to get $x$ .

$x = - \frac{4}{2}$

Simplify the right side of the equation.

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Divide $4$ by $2$ to get $2$ .

$x = - 1 \cdot 2$

Multiply $- 1$ by $2$ to get $- 2$ .

$x = - 2$

Solve the original function $f (x) = x^{2} + 4 x - 8$ at $x = - 2$ .

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Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {(- 2)}^{2} + 4 (- 2) - 8$

Simplify the result.

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Simplify each term.

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Raise $- 2$ to the power of $2$ to get $4$ .

$f (- 2) = 4 + 4 (- 2) - 8$

Multiply $4$ by $- 2$ to get $- 8$ .

$f (- 2) = 4 - 8 - 8$

Simplify by subtracting numbers.

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Subtract $8$ from $4$ to get $- 4$ .

$f (- 2) = - 4 - 8$

Subtract $8$ from $- 4$ to get $- 12$ .

$f (- 2) = - 12$

The final answer is $- 12$ .

$- 12$

The horizontal tangent lines on function $f (x) = x^{2} + 4 x - 8$ are $y = - 12$ .

$y = - 12$

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