Calculus Examples

$f (x) = (- 5 x^{2} + 5 x - 2) (- 2 x + 3)$ , $x = 0$

Step 1

Find the corresponding $y$ -value to $x = 0$ .

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Step 1.1

Substitute $0$ in for $x$ .

$y = (- 5 {(0)}^{2} + 5 (0) - 2) (- 2 \cdot 0 + 3)$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = (- 5 \cdot 0^{2} + 5 (0) - 2) (- 2 \cdot 0 + 3)$

Step 1.2.2

Remove parentheses.

$y = (- 5 {(0)}^{2} + 5 (0) - 2) (- 2 \cdot 0 + 3)$

Step 1.2.3

Simplify $(- 5 {(0)}^{2} + 5 (0) - 2) (- 2 \cdot 0 + 3)$ .

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Raising $0$ to any positive power yields $0$ .

$y = (- 5 \cdot 0 + 5 (0) - 2) (- 2 \cdot 0 + 3)$

Step 1.2.3.1.2

Multiply $- 5$ by $0$ .

$y = (0 + 5 (0) - 2) (- 2 \cdot 0 + 3)$

Step 1.2.3.1.3

Multiply $5$ by $0$ .

$y = (0 + 0 - 2) (- 2 \cdot 0 + 3)$

Step 1.2.3.2

Simplify the expression.

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Step 1.2.3.2.1

Add $0$ and $0$ .

$y = (0 - 2) (- 2 \cdot 0 + 3)$

Step 1.2.3.2.2

Subtract $2$ from $0$ .

$y = - 2 (- 2 \cdot 0 + 3)$

Step 1.2.3.2.3

Multiply $- 2$ by $0$ .

$y = - 2 (0 + 3)$

Step 1.2.3.2.4

Add $0$ and $3$ .

$y = - 2 \cdot 3$

Step 1.2.3.2.5

Multiply $- 2$ by $3$ .

$y = - 6$

Step 2

Find the first derivative and evaluate at $x = 0$ and $y = - 6$ to find the slope of the tangent line.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 5 x^{2} + 5 x - 2$ and $g (x) = - 2 x + 3$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

$(- 5 x^{2} + 5 x - 2) (- 2 \cdot 1 + \frac{d}{d x} [3]) + (- 2 x + 3) \frac{d}{d x} [- 5 x^{2} + 5 x - 2]$

Step 2.2.4

Multiply $- 2$ by $1$ .

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

Step 2.2.5

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

$(- 5 x^{2} + 5 x - 2) (- 2 + 0) + (- 2 x + 3) \frac{d}{d x} [- 5 x^{2} + 5 x - 2]$

Step 2.2.6

Simplify the expression.

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Step 2.2.6.1

Add $- 2$ and $0$ .

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

Step 2.2.6.2

Move $- 2$ to the left of $- 5 x^{2} + 5 x - 2$ .

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

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

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

Step 2.2.10

Multiply $2$ by $- 5$ .

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 2])$

Step 2.2.11

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

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + 5 \frac{d}{d x} [x] + \frac{d}{d x} [- 2])$

Step 2.2.12

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

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + 5 \cdot 1 + \frac{d}{d x} [- 2])$

Step 2.2.13

Multiply $5$ by $1$ .

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + 5 + \frac{d}{d x} [- 2])$

Step 2.2.14

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

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + 5 + 0)$

Step 2.2.15

Add $- 10 x + 5$ and $0$ .

$- 2 (- 5 x^{2} + 5 x - 2) + (- 2 x + 3) (- 10 x + 5)$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$- 2 (- 5 x^{2}) - 2 (5 x) - 2 \cdot - 2 + (- 2 x + 3) (- 10 x + 5)$

Step 2.3.2

Apply the distributive property.

$- 2 (- 5 x^{2}) - 2 (5 x) - 2 \cdot - 2 + (- 2 x + 3) (- 10 x) + (- 2 x + 3) \cdot 5$

Step 2.3.3

Apply the distributive property.

$- 2 (- 5 x^{2}) - 2 (5 x) - 2 \cdot - 2 - 2 x (- 10 x) + 3 (- 10 x) + (- 2 x + 3) \cdot 5$

Step 2.3.4

Apply the distributive property.

$- 2 (- 5 x^{2}) - 2 (5 x) - 2 \cdot - 2 - 2 x (- 10 x) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5

Combine terms.

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Step 2.3.5.1

Multiply $- 5$ by $- 2$ .

$10 x^{2} - 2 (5 x) - 2 \cdot - 2 - 2 x (- 10 x) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.2

Multiply $5$ by $- 2$ .

$10 x^{2} - 10 x - 2 \cdot - 2 - 2 x (- 10 x) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.3

Multiply $- 2$ by $- 2$ .

$10 x^{2} - 10 x + 4 - 2 x (- 10 x) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.4

Multiply $- 10$ by $- 2$ .

$10 x^{2} - 10 x + 4 + 20 x \cdot x + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.5

Raise $x$ to the power of $1$ .

$10 x^{2} - 10 x + 4 + 20 (x^{1} x) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.6

Raise $x$ to the power of $1$ .

$10 x^{2} - 10 x + 4 + 20 (x^{1} x^{1}) + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$10 x^{2} - 10 x + 4 + 20 x^{1 + 1} + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.8

Add $1$ and $1$ .

$10 x^{2} - 10 x + 4 + 20 x^{2} + 3 (- 10 x) - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.9

Multiply $- 10$ by $3$ .

$10 x^{2} - 10 x + 4 + 20 x^{2} - 30 x - 2 x \cdot 5 + 3 \cdot 5$

Step 2.3.5.10

Multiply $5$ by $- 2$ .

$10 x^{2} - 10 x + 4 + 20 x^{2} - 30 x - 10 x + 3 \cdot 5$

Step 2.3.5.11

Multiply $3$ by $5$ .

$10 x^{2} - 10 x + 4 + 20 x^{2} - 30 x - 10 x + 15$

Step 2.3.5.12

Subtract $10 x$ from $- 30 x$ .

$10 x^{2} - 10 x + 4 + 20 x^{2} - 40 x + 15$

Step 2.3.5.13

Add $10 x^{2}$ and $20 x^{2}$ .

$30 x^{2} - 10 x + 4 - 40 x + 15$

Step 2.3.5.14

Subtract $40 x$ from $- 10 x$ .

$30 x^{2} - 50 x + 4 + 15$

Step 2.3.5.15

Add $4$ and $15$ .

$30 x^{2} - 50 x + 19$

Step 2.4

Evaluate the derivative at $x = 0$ .

$30 {(0)}^{2} - 50 \cdot 0 + 19$

Step 2.5

Simplify.

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Step 2.5.1

Simplify each term.

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Step 2.5.1.1

Raising $0$ to any positive power yields $0$ .

$30 \cdot 0 - 50 \cdot 0 + 19$

Step 2.5.1.2

Multiply $30$ by $0$ .

$0 - 50 \cdot 0 + 19$

Step 2.5.1.3

Multiply $- 50$ by $0$ .

$0 + 0 + 19$

Step 2.5.2

Simplify by adding numbers.

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Step 2.5.2.1

Add $0$ and $0$ .

$0 + 19$

Step 2.5.2.2

Add $0$ and $19$ .

$19$

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 $19$ and a given point $(0, - 6)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 6) = 19 \cdot (x - (0))$

Step 3.2

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

$y + 6 = 19 \cdot (x + 0)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Add $x$ and $0$ .

$y + 6 = 19 x$

Step 3.3.2

Subtract $6$ from both sides of the equation.

$y = 19 x - 6$

Step 4