Find the Standard Form of the Parabola f(x)=3x^2-x+1

$f (x) = 3 x^{2} - x + 1$

Step 1

Write $f (x) = 3 x^{2} - x + 1$ as an equation.

$y = 3 x^{2} - x + 1$

Step 2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - x + 1 = y$

Step 3

Subtract $1$ from both sides of the equation.

$3 x^{2} - x = y - 1$

Step 4

Complete the square.

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

$a = 3$

$b = - 1$

$c = 0 = y - 1$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y - 1$

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

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

$d = \frac{- 1}{2 \cdot 3}$

Simplify the right side.

Multiply $2$ by $3$ .

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

Move the negative in front of the fraction.

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

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

$e = 0 - \frac{{(- 1)}^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{1}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 0 - \frac{1}{12}$

Subtract $\frac{1}{12}$ from $0$ .

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

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x - \frac{1}{6})}^{2} - \frac{1}{12}$ .

$3 {(x - \frac{1}{6})}^{2} - \frac{1}{12} = y - 1$

Step 5

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

Add $\frac{1}{12}$ to both sides of the equation.

$3 {(x - \frac{1}{6})}^{2} = y - 1 + \frac{1}{12}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$3 {(x - \frac{1}{6})}^{2} = y - 1 \cdot \frac{12}{12} + \frac{1}{12}$

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

$3 {(x - \frac{1}{6})}^{2} = y + \frac{- 1 \cdot 12}{12} + \frac{1}{12}$

Combine the numerators over the common denominator.

$3 {(x - \frac{1}{6})}^{2} = y + \frac{- 1 \cdot 12 + 1}{12}$

Simplify the numerator.

Multiply $- 1$ by $12$ .

$3 {(x - \frac{1}{6})}^{2} = y + \frac{- 12 + 1}{12}$

Add $- 12$ and $1$ .

$3 {(x - \frac{1}{6})}^{2} = y + \frac{- 11}{12}$

Move the negative in front of the fraction.

$3 {(x - \frac{1}{6})}^{2} = y - \frac{11}{12}$

Step 6

Divide each term in $3 {(x - \frac{1}{6})}^{2} = y - \frac{11}{12}$ by $3$ and simplify.

Divide each term in $3 {(x - \frac{1}{6})}^{2} = y - \frac{11}{12}$ by $3$ .

$\frac{3 {(x - \frac{1}{6})}^{2}}{3} = \frac{y}{3} + \frac{- \frac{11}{12}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 {(x - \frac{1}{6})}^{2}}{3} = \frac{y}{3} + \frac{- \frac{11}{12}}{3}$

Divide ${(x - \frac{1}{6})}^{2}$ by $1$ .

${(x - \frac{1}{6})}^{2} = \frac{y}{3} + \frac{- \frac{11}{12}}{3}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

${(x - \frac{1}{6})}^{2} = \frac{y}{3} - \frac{11}{12} \cdot \frac{1}{3}$

Multiply $- \frac{11}{12} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{3}$ by $\frac{11}{12}$ .

${(x - \frac{1}{6})}^{2} = \frac{y}{3} - \frac{11}{3 \cdot 12}$

Multiply $3$ by $12$ .

${(x - \frac{1}{6})}^{2} = \frac{y}{3} - \frac{11}{36}$

Step 7

Factor.

To write $\frac{y}{3}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

${(x - \frac{1}{6})}^{2} = \frac{y}{3} \cdot \frac{12}{12} - \frac{11}{36}$

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{y}{3}$ by $\frac{12}{12}$ .

${(x - \frac{1}{6})}^{2} = \frac{y \cdot 12}{3 \cdot 12} - \frac{11}{36}$

Multiply $3$ by $12$ .

${(x - \frac{1}{6})}^{2} = \frac{y \cdot 12}{36} - \frac{11}{36}$

Combine the numerators over the common denominator.

${(x - \frac{1}{6})}^{2} = \frac{y \cdot 12 - 11}{36}$

Move $12$ to the left of $y$ .

${(x - \frac{1}{6})}^{2} = \frac{12 y - 11}{36}$

Reorder terms.

${(x - \frac{1}{6})}^{2} = \frac{1}{36} (12 y - 11)$

Find the Standard Form of the Parabola f(x)=3x^2-x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.