Find the Vertex 12y=(x-1)^2-48

$12 y = {(x - 1)}^{2} - 48$

Step 1

Rewrite the equation in vertex form.

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

Divide each term in $12 y = {(x - 1)}^{2} - 48$ by $12$ and simplify.

Divide each term in $12 y = {(x - 1)}^{2} - 48$ by $12$ .

$\frac{12 y}{12} = \frac{{(x - 1)}^{2}}{12} + \frac{- 48}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 y}{12} = \frac{{(x - 1)}^{2}}{12} + \frac{- 48}{12}$

Divide $y$ by $1$ .

$y = \frac{{(x - 1)}^{2}}{12} + \frac{- 48}{12}$

Simplify the right side.

Divide $- 48$ by $12$ .

$y = \frac{{(x - 1)}^{2}}{12} - 4$

Reorder terms.

$y = \frac{1}{12} \cdot {(x - 1)}^{2} - 4$

Complete the square for $\frac{1}{12} \cdot {(x - 1)}^{2} - 4$ .

Simplify the expression.

Simplify each term.

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

$\frac{1}{12} \cdot ((x - 1) (x - 1)) - 4$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$\frac{1}{12} \cdot (x (x - 1) - 1 (x - 1)) - 4$

Apply the distributive property.

$\frac{1}{12} \cdot (x \cdot x + x \cdot - 1 - 1 (x - 1)) - 4$

Apply the distributive property.

$\frac{1}{12} \cdot (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) - 4$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\frac{1}{12} \cdot (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) - 4$

Move $- 1$ to the left of $x$ .

$\frac{1}{12} \cdot (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) - 4$

Rewrite $- 1 x$ as $- x$ .

$\frac{1}{12} \cdot (x^{2} - x - 1 x - 1 \cdot - 1) - 4$

Rewrite $- 1 x$ as $- x$ .

$\frac{1}{12} \cdot (x^{2} - x - x - 1 \cdot - 1) - 4$

Multiply $- 1$ by $- 1$ .

$\frac{1}{12} \cdot (x^{2} - x - x + 1) - 4$

Subtract $x$ from $- x$ .

$\frac{1}{12} \cdot (x^{2} - 2 x + 1) - 4$

Apply the distributive property.

$\frac{1}{12} x^{2} + \frac{1}{12} (- 2 x) + \frac{1}{12} \cdot 1 - 4$

Simplify.

Combine $\frac{1}{12}$ and $x^{2}$ .

$\frac{x^{2}}{12} + \frac{1}{12} (- 2 x) + \frac{1}{12} \cdot 1 - 4$

Cancel the common factor of $2$ .

Factor $2$ out of $12$ .

$\frac{x^{2}}{12} + \frac{1}{2 (6)} (- 2 x) + \frac{1}{12} \cdot 1 - 4$

Factor $2$ out of $- 2 x$ .

$\frac{x^{2}}{12} + \frac{1}{2 (6)} (2 (- x)) + \frac{1}{12} \cdot 1 - 4$

Cancel the common factor.

$\frac{x^{2}}{12} + \frac{1}{2 \cdot 6} (2 (- x)) + \frac{1}{12} \cdot 1 - 4$

Rewrite the expression.

$\frac{x^{2}}{12} + \frac{1}{6} (- x) + \frac{1}{12} \cdot 1 - 4$

Combine $\frac{1}{6}$ and $x$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1}{12} \cdot 1 - 4$

Multiply $\frac{1}{12}$ by $1$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1}{12} - 4$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1}{12} - 4 \cdot \frac{12}{12}$

Combine $- 4$ and $\frac{12}{12}$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1}{12} + \frac{- 4 \cdot 12}{12}$

Combine the numerators over the common denominator.

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1 - 4 \cdot 12}{12}$

Simplify the numerator.

Multiply $- 4$ by $12$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{1 - 48}{12}$

Subtract $48$ from $1$ .

$\frac{x^{2}}{12} - \frac{x}{6} + \frac{- 47}{12}$

Move the negative in front of the fraction.

$\frac{x^{2}}{12} - \frac{x}{6} - \frac{47}{12}$

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

$a = \frac{1}{12}$

$b = - \frac{1}{6}$

$c = - \frac{47}{12}$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{- \frac{1}{6}}{2 (\frac{1}{12})}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$d = - \frac{1}{6} \cdot \frac{1}{2 (\frac{1}{12})}$

Combine $2$ and $\frac{1}{12}$ .

$d = - \frac{1}{6} \cdot \frac{1}{\frac{2}{12}}$

Cancel the common factor of $2$ and $12$ .

Factor $2$ out of $2$ .

$d = - \frac{1}{6} \cdot \frac{1}{\frac{2 (1)}{12}}$

Cancel the common factors.

Factor $2$ out of $12$ .

$d = - \frac{1}{6} \cdot \frac{1}{\frac{2 \cdot 1}{2 \cdot 6}}$

Cancel the common factor.

$d = - \frac{1}{6} \cdot \frac{1}{\frac{2 \cdot 1}{2 \cdot 6}}$

Rewrite the expression.

$d = - \frac{1}{6} \cdot \frac{1}{\frac{1}{6}}$

Multiply the numerator by the reciprocal of the denominator.

$d = - \frac{1}{6} (1 \cdot 6)$

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{1}{6}$ into the numerator.

$d = \frac{- 1}{6} (1 \cdot 6)$

Factor $6$ out of $1 \cdot 6$ .

$d = \frac{- 1}{6} (6 \cdot 1)$

Cancel the common factor.

$d = \frac{- 1}{6} (6 \cdot 1)$

Rewrite the expression.

$d = - 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = - \frac{47}{12} - \frac{{(- \frac{1}{6})}^{2}}{4 (\frac{1}{12})}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $- \frac{1}{6}$ .

$e = - \frac{47}{12} - \frac{{(- 1)}^{2} {(\frac{1}{6})}^{2}}{4 (\frac{1}{12})}$

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

$e = - \frac{47}{12} - \frac{1 {(\frac{1}{6})}^{2}}{4 (\frac{1}{12})}$

Apply the product rule to $\frac{1}{6}$ .

$e = - \frac{47}{12} - \frac{1 \frac{1^{2}}{6^{2}}}{4 (\frac{1}{12})}$

One to any power is one.

$e = - \frac{47}{12} - \frac{1 \frac{1}{6^{2}}}{4 (\frac{1}{12})}$

Raise $6$ to the power of $2$ .

$e = - \frac{47}{12} - \frac{1 (\frac{1}{36})}{4 (\frac{1}{12})}$

Multiply $\frac{1}{36}$ by $1$ .

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{4 (\frac{1}{12})}$

Combine $4$ and $\frac{1}{12}$ .

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{\frac{4}{12}}$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4$ .

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{\frac{4 (1)}{12}}$

Cancel the common factors.

Factor $4$ out of $12$ .

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Cancel the common factor.

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Rewrite the expression.

$e = - \frac{47}{12} - \frac{\frac{1}{36}}{\frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$e = - \frac{47}{12} - (\frac{1}{36} \cdot 3)$

Cancel the common factor of $3$ .

Factor $3$ out of $36$ .

$e = - \frac{47}{12} - (\frac{1}{3 (12)} \cdot 3)$

Cancel the common factor.

$e = - \frac{47}{12} - (\frac{1}{3 \cdot 12} \cdot 3)$

Rewrite the expression.

$e = - \frac{47}{12} - \frac{1}{12}$

Combine the numerators over the common denominator.

$e = \frac{- 47 - 1}{12}$

Subtract $1$ from $- 47$ .

$e = \frac{- 48}{12}$

Divide $- 48$ by $12$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $\frac{1}{12} {(x - 1)}^{2} - 4$ .

$\frac{1}{12} {(x - 1)}^{2} - 4$

Set $y$ equal to the new right side.

$y = \frac{1}{12} \cdot {(x - 1)}^{2} - 4$

Step 2

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

$a = \frac{1}{12}$

$h = 1$

$k = - 4$

Step 3

Find the vertex $(h, k)$ .

$(1, - 4)$

Step 4

Find the Vertex 12y=(x-1)^2-48

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