Find the Standard Form of the Parabola 3x^2+30x+y+79=0

$3 x^{2} + 30 x + y + 79 = 0$

Step 1

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

Subtract $y$ from both sides of the equation.

$3 x^{2} + 30 x + 79 = - y$

Subtract $79$ from both sides of the equation.

$3 x^{2} + 30 x = - y - 79$

Step 2

Complete the square.

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

$a = 3$

$b = 30$

$c = 0 = - y - 79$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - y - 79$

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

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

$d = \frac{30}{2 \cdot 3}$

Simplify the right side.

Cancel the common factor of $30$ and $2$ .

Factor $2$ out of $30$ .

$d = \frac{2 \cdot 15}{2 \cdot 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 3$ .

$d = \frac{2 \cdot 15}{2 (3)}$

Cancel the common factor.

$d = \frac{2 \cdot 15}{2 \cdot 3}$

Rewrite the expression.

$d = \frac{15}{3}$

Cancel the common factor of $15$ and $3$ .

Factor $3$ out of $15$ .

$d = \frac{3 \cdot 5}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$d = \frac{3 \cdot 5}{3 (1)}$

Cancel the common factor.

$d = \frac{3 \cdot 5}{3 \cdot 1}$

Rewrite the expression.

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

Divide $5$ by $1$ .

$d = 5$

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

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

$e = 0 - \frac{30^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raise $30$ to the power of $2$ .

$e = 0 - \frac{900}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 0 - \frac{900}{12}$

Divide $900$ by $12$ .

$e = 0 - 1 \cdot 75$

Multiply $- 1$ by $75$ .

$e = 0 - 75$

Subtract $75$ from $0$ .

$e = - 75$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x + 5)}^{2} - 75$ .

$3 {(x + 5)}^{2} - 75 = - y - 79$

Step 3

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

Add $75$ to both sides of the equation.

$3 {(x + 5)}^{2} = - y - 79 + 75$

Add $- 79$ and $75$ .

$3 {(x + 5)}^{2} = - y - 4$

Step 4

Divide each term in $3 {(x + 5)}^{2} = - y - 4$ by $3$ and simplify.

Divide each term in $3 {(x + 5)}^{2} = - y - 4$ by $3$ .

$\frac{3 {(x + 5)}^{2}}{3} = \frac{- y}{3} + \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 {(x + 5)}^{2}}{3} = \frac{- y}{3} + \frac{- 4}{3}$

Divide ${(x + 5)}^{2}$ by $1$ .

${(x + 5)}^{2} = \frac{- y}{3} + \frac{- 4}{3}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

${(x + 5)}^{2} = - \frac{y}{3} + \frac{- 4}{3}$

Move the negative in front of the fraction.

${(x + 5)}^{2} = - \frac{y}{3} - \frac{4}{3}$

Step 5

Factor.

Factor $- 1$ out of $- \frac{y}{3} - \frac{4}{3}$ .

Factor $- 1$ out of $- \frac{y}{3}$ .

${(x + 5)}^{2} = - (\frac{y}{3}) - \frac{4}{3}$

Factor $- 1$ out of $- \frac{4}{3}$ .

${(x + 5)}^{2} = - (\frac{y}{3}) - (\frac{4}{3})$

Factor $- 1$ out of $- (\frac{y}{3}) - (\frac{4}{3})$ .

${(x + 5)}^{2} = - (\frac{y}{3} + \frac{4}{3})$

Combine the numerators over the common denominator.

${(x + 5)}^{2} = - \frac{y + 4}{3}$

Reorder terms.

${(x + 5)}^{2} = - (\frac{1}{3} (y + 4))$

Remove parentheses.

${(x + 5)}^{2} = - \frac{1}{3} (y + 4)$

Find the Standard Form of the Parabola 3x^2+30x+y+79=0

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