Calculus Examples

$f (x) = 2 x^{2} + x - 3$ , $x = 4$

Step 1

Find the corresponding $y$ -value to $x = 4$ .

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Step 1.1

Substitute $4$ in for $x$ .

$y = 2 {(4)}^{2} + 4 - 3$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = 2 {(4)}^{2} + 4 - 3$

Step 1.2.2

Remove parentheses.

$y = 2 \cdot 4^{2} + 4 - 3$

Step 1.2.3

Remove parentheses.

$y = 2 {(4)}^{2} + 4 - 3$

Step 1.2.4

Simplify $2 {(4)}^{2} + 4 - 3$ .

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Step 1.2.4.1

Simplify each term.

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Step 1.2.4.1.1

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

$y = 2 \cdot 16 + 4 - 3$

Step 1.2.4.1.2

Multiply $2$ by $16$ .

$y = 32 + 4 - 3$

Step 1.2.4.2

Simplify by adding and subtracting.

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Step 1.2.4.2.1

Add $32$ and $4$ .

$y = 36 - 3$

Step 1.2.4.2.2

Subtract $3$ from $36$ .

$y = 33$

Step 2

Find the first derivative and evaluate at $x = 4$ and $y = 33$ to find the slope of the tangent line.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 2.3

Differentiate.

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Step 2.3.1

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

$4 x + 1 + \frac{d}{d x} [- 3]$

Step 2.3.2

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$4 x + 1 + 0$

Step 2.3.3

Add $4 x + 1$ and $0$ .

$4 x + 1$

Step 2.4

Evaluate the derivative at $x = 4$ .

$4 (4) + 1$

Step 2.5

Simplify.

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Step 2.5.1

Multiply $4$ by $4$ .

$16 + 1$

Step 2.5.2

Add $16$ and $1$ .

$17$

Step 3

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

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Step 3.1

Use the slope $17$ and a given point $(4, 33)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (33) = 17 \cdot (x - (4))$

Step 3.2

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

$y - 33 = 17 \cdot (x - 4)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $17 \cdot (x - 4)$ .

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Step 3.3.1.1

Rewrite.

$y - 33 = 0 + 0 + 17 \cdot (x - 4)$

Step 3.3.1.2

Simplify by adding zeros.

$y - 33 = 17 \cdot (x - 4)$

Step 3.3.1.3

Apply the distributive property.

$y - 33 = 17 x + 17 \cdot - 4$

Step 3.3.1.4

Multiply $17$ by $- 4$ .

$y - 33 = 17 x - 68$

Step 3.3.2

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

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Step 3.3.2.1

Add $33$ to both sides of the equation.

$y = 17 x - 68 + 33$

Step 3.3.2.2

Add $- 68$ and $33$ .

$y = 17 x - 35$

Step 4