Precalculus Examples

$5 x^{2} + 40 x - y + 78 = 0$

Step 1

Rewrite the equation in vertex form.

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Step 1.1

Isolate $y$ to the left side of the equation.

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Step 1.1.1

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

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Step 1.1.1.1

Subtract $5 x^{2}$ from both sides of the equation.

$40 x - y + 78 = - 5 x^{2}$

Step 1.1.1.2

Subtract $40 x$ from both sides of the equation.

$- y + 78 = - 5 x^{2} - 40 x$

Step 1.1.1.3

Subtract $78$ from both sides of the equation.

$- y = - 5 x^{2} - 40 x - 78$

Step 1.1.2

Divide each term in $- y = - 5 x^{2} - 40 x - 78$ by $- 1$ and simplify.

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Step 1.1.2.1

Divide each term in $- y = - 5 x^{2} - 40 x - 78$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 5 x^{2}}{- 1} + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 5 x^{2}}{- 1} + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.2.2

Divide $y$ by $1$ .

$y = \frac{- 5 x^{2}}{- 1} + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Move the negative one from the denominator of $\frac{- 5 x^{2}}{- 1}$ .

$y = - 1 \cdot (- 5 x^{2}) + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.3.1.2

Rewrite $- 1 \cdot (- 5 x^{2})$ as $- (- 5 x^{2})$ .

$y = - (- 5 x^{2}) + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.3.1.3

Multiply $- 5$ by $- 1$ .

$y = 5 x^{2} + \frac{- 40 x}{- 1} + \frac{- 78}{- 1}$

Step 1.1.2.3.1.4

Move the negative one from the denominator of $\frac{- 40 x}{- 1}$ .

$y = 5 x^{2} - 1 \cdot (- 40 x) + \frac{- 78}{- 1}$

Step 1.1.2.3.1.5

Rewrite $- 1 \cdot (- 40 x)$ as $- (- 40 x)$ .

$y = 5 x^{2} - (- 40 x) + \frac{- 78}{- 1}$

Step 1.1.2.3.1.6

Multiply $- 40$ by $- 1$ .

$y = 5 x^{2} + 40 x + \frac{- 78}{- 1}$

Step 1.1.2.3.1.7

Divide $- 78$ by $- 1$ .

$y = 5 x^{2} + 40 x + 78$

Step 1.2

Complete the square for $5 x^{2} + 40 x + 78$ .

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Step 1.2.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 5$

$b = 40$

$c = 78$

Step 1.2.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.2.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.2.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{40}{2 \cdot 5}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $40$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $40$ .

$d = \frac{2 \cdot 20}{2 \cdot 5}$

Step 1.2.3.2.1.2

Cancel the common factors.

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Step 1.2.3.2.1.2.1

Factor $2$ out of $2 \cdot 5$ .

$d = \frac{2 \cdot 20}{2 (5)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 20}{2 \cdot 5}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{20}{5}$

Step 1.2.3.2.2

Cancel the common factor of $20$ and $5$ .

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Step 1.2.3.2.2.1

Factor $5$ out of $20$ .

$d = \frac{5 \cdot 4}{5}$

Step 1.2.3.2.2.2

Cancel the common factors.

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Step 1.2.3.2.2.2.1

Factor $5$ out of $5$ .

$d = \frac{5 \cdot 4}{5 (1)}$

Step 1.2.3.2.2.2.2

Cancel the common factor.

$d = \frac{5 \cdot 4}{5 \cdot 1}$

Step 1.2.3.2.2.2.3

Rewrite the expression.

$d = \frac{4}{1}$

Step 1.2.3.2.2.2.4

Divide $4$ by $1$ .

$d = 4$

Step 1.2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 78 - \frac{40^{2}}{4 \cdot 5}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $40$ to the power of $2$ .

$e = 78 - \frac{1600}{4 \cdot 5}$

Step 1.2.4.2.1.2

Multiply $4$ by $5$ .

$e = 78 - \frac{1600}{20}$

Step 1.2.4.2.1.3

Divide $1600$ by $20$ .

$e = 78 - 1 \cdot 80$

Step 1.2.4.2.1.4

Multiply $- 1$ by $80$ .

$e = 78 - 80$

Step 1.2.4.2.2

Subtract $80$ from $78$ .

$e = - 2$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $5 {(x + 4)}^{2} - 2$ .

$5 {(x + 4)}^{2} - 2$

Step 1.3

Set $y$ equal to the new right side.

$y = 5 {(x + 4)}^{2} - 2$

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 5$

$h = - 4$

$k = - 2$

Step 3

Find the vertex $(h, k)$ .

$(- 4, - 2)$

Step 4