Calculus Examples

$f (x) = 8 x^{2}$ , $16 x + y + 6 = 0$

Step 1

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

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Step 1.1

Subtract $16 x$ from both sides of the equation.

$y + 6 = - 16 x$

Step 1.2

Subtract $6$ from both sides of the equation.

$y = - 16 x - 6$

Step 2

Use the slope-intercept form to find the slope.

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Step 2.1

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Step 2.2

Using the slope-intercept form, the slope is $- 16$ .

$m = - 16$

Step 3

Find the derivative.

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Step 3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

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

Step 3.2

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

$8 (2 x)$

Step 3.3

Multiply $2$ by $8$ .

$16 x$

Step 4

The first derivative of a function represents the slope at every point of that function. In this case, the derivative of $f (x) = 8 x^{2}$ is $16 x$ and the slope of the given line $y = - 16 x - 6$ is $m = - 16$ . To find the point on $f (x) = 8 x^{2}$ where the slope of the tangent line is the same as the slope of the given line $y = - 16 x - 6$ , substitute the value of the slope of the given line $- 16$ for the value of $16 x$ .

$- 16 = 16 x$

Step 5

Solve $- 16 = 16 x$ for $x$ to find the x-coordinate of the point where the tangent line is parallel to the given line $y = - 16 x - 6$ . In this case, the x-coordinate is $x = - 1$ .

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Step 5.1

Rewrite the equation as $16 x = - 16$ .

$16 x = - 16$

Step 5.2

Divide each term in $16 x = - 16$ by $16$ and simplify.

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Step 5.2.1

Divide each term in $16 x = - 16$ by $16$ .

$\frac{16 x}{16} = \frac{- 16}{16}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $16$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{16 x}{16} = \frac{- 16}{16}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 16}{16}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $- 16$ by $16$ .

$x = - 1$

Step 6

Substitute $- 1$ in $f (x) = 8 x^{2}$ to get the y-coordinate of the point where the tangent line is parallel to the given line $y = - 16 x - 6$ . In this case, the y-coordinate is $8$ .

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 8 {(- 1)}^{2}$

Step 6.2

Simplify the result.

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Step 6.2.1

Raise $- 1$ to the power of $2$ .

$f (- 1) = 8 \cdot 1$

Step 6.2.2

Multiply $8$ by $1$ .

$f (- 1) = 8$

Step 6.2.3

The final answer is $8$ .

$8$

Step 7

The point on $f (x) = 8 x^{2}$ where the slope of the tangent line is the same as the slope of the given line $y = - 16 x - 6$ has x-coordinate of $- 1$ and y-coordinate of $8$ . The slope of the tangent line is the same as the slope of $y = - 16 x - 6$ , which is $m = - 16$ .

$(- 1, 8), m = - 16$

Step 8

The tangent line on $(- 1, 8)$ where the slope is $m = - 16$ .

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Step 8.1

Find the value of $b$ using the formula for the equation of a line.

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Step 8.1.1

Use the formula for the equation of a line to find $b$ .

$y = m x + b$

Step 8.1.2

Substitute the value of $m$ into the equation.

$y = (- 16) \cdot x + b$

Step 8.1.3

Substitute the value of $x$ into the equation.

$y = (- 16) \cdot (- 1) + b$

Step 8.1.4

Substitute the value of $y$ into the equation.

$8 = (- 16) \cdot (- 1) + b$

Step 8.1.5

Find the value of $b$ .

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Step 8.1.5.1

Rewrite the equation as $(- 16) \cdot (- 1) + b = 8$ .

$(- 16) \cdot (- 1) + b = 8$

Step 8.1.5.2

Multiply $- 16$ by $- 1$ .

$16 + b = 8$

Step 8.1.5.3

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

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Step 8.1.5.3.1

Subtract $16$ from both sides of the equation.

$b = 8 - 16$

Step 8.1.5.3.2

Subtract $16$ from $8$ .

$b = - 8$

Step 8.2

Now that the values of $m$ (slope) and $b$ (y-intercept) are known, substitute them into $y = m x + b$ to find the equation of the line.

$y = - 16 x - 8$

Step 9