Graph y=3x^2-2x+1

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

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $3 x^{2} - 2 x + 1$ .

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

$a = 3$

$b = - 2$

$c = 1$

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 $d = \frac{b}{2 a}$ .

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

Simplify the right side.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $3$ .

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

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

Factor $4$ out of $4$ .

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

Cancel the common factors.

Factor $4$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

$e = 1 - \frac{1}{3}$

Write $1$ as a fraction with a common denominator.

$e = \frac{3}{3} - \frac{1}{3}$

Combine the numerators over the common denominator.

$e = \frac{3 - 1}{3}$

Subtract $1$ from $3$ .

$e = \frac{2}{3}$

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

$3 {(x - \frac{1}{3})}^{2} + \frac{2}{3}$

Set $y$ equal to the new right side.

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

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

$a = 3$

$h = \frac{1}{3}$

$k = \frac{2}{3}$

Since the value of $a$ is positive, the parabola opens up.

Opens Up

Find the vertex $(h, k)$ .

$(\frac{1}{3}, \frac{2}{3})$

Find $p$ , the distance from the vertex to the focus.

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot 3}$

Multiply $4$ by $3$ .

$\frac{1}{12}$

Find the focus.

The focus of a parabola can be found by adding $p$ to the y-coordinate $k$ if the parabola opens up or down.

$(h, k + p)$

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(\frac{1}{3}, \frac{3}{4})$

Find the axis of symmetry by finding the line that passes through the vertex and the focus.

$x = \frac{1}{3}$

Find the directrix.

The directrix of a parabola is the horizontal line found by subtracting $p$ from the y-coordinate $k$ of the vertex if the parabola opens up or down.

$y = k - p$

Substitute the known values of $p$ and $k$ into the formula and simplify.

$y = \frac{7}{12}$

Use the properties of the parabola to analyze and graph the parabola.

Direction: Opens Up

Vertex: $(\frac{1}{3}, \frac{2}{3})$

Focus: $(\frac{1}{3}, \frac{3}{4})$

Axis of Symmetry: $x = \frac{1}{3}$

Directrix: $y = \frac{7}{12}$

Direction: Opens Up

Vertex: $(\frac{1}{3}, \frac{2}{3})$

Focus: $(\frac{1}{3}, \frac{3}{4})$

Axis of Symmetry: $x = \frac{1}{3}$

Directrix: $y = \frac{7}{12}$

Step 2

Select a few $x$ values, and plug them into the equation to find the corresponding $y$ values. The $x$ values should be selected around the vertex.

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

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

Simplify the result.

Simplify each term.

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

$f (- 1) = 3 \cdot 1 - 2 \cdot - 1 + 1$

Multiply $3$ by $1$ .

$f (- 1) = 3 - 2 \cdot - 1 + 1$

Multiply $- 2$ by $- 1$ .

$f (- 1) = 3 + 2 + 1$

Simplify by adding numbers.

Add $3$ and $2$ .

$f (- 1) = 5 + 1$

Add $5$ and $1$ .

$f (- 1) = 6$

The final answer is $6$ .

$6$

The $y$ value at $x = - 1$ is $6$ .

$y = 6$

Replace the variable $x$ with $- 2$ in the expression.

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

Simplify the result.

Simplify each term.

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

$f (- 2) = 3 \cdot 4 - 2 \cdot - 2 + 1$

Multiply $3$ by $4$ .

$f (- 2) = 12 - 2 \cdot - 2 + 1$

Multiply $- 2$ by $- 2$ .

$f (- 2) = 12 + 4 + 1$

Simplify by adding numbers.

Add $12$ and $4$ .

$f (- 2) = 16 + 1$

Add $16$ and $1$ .

$f (- 2) = 17$

The final answer is $17$ .

$17$

The $y$ value at $x = - 2$ is $17$ .

$y = 17$

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

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

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 3 \cdot 1 - 2 \cdot 1 + 1$

Multiply $3$ by $1$ .

$f (1) = 3 - 2 \cdot 1 + 1$

Multiply $- 2$ by $1$ .

$f (1) = 3 - 2 + 1$

Simplify by adding and subtracting.

Subtract $2$ from $3$ .

$f (1) = 1 + 1$

Add $1$ and $1$ .

$f (1) = 2$

The final answer is $2$ .

$2$

The $y$ value at $x = 1$ is $2$ .

$y = 2$

Replace the variable $x$ with $2$ in the expression.

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

Simplify the result.

Simplify each term.

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

$f (2) = 3 \cdot 4 - 2 \cdot 2 + 1$

Multiply $3$ by $4$ .

$f (2) = 12 - 2 \cdot 2 + 1$

Multiply $- 2$ by $2$ .

$f (2) = 12 - 4 + 1$

Simplify by adding and subtracting.

Subtract $4$ from $12$ .

$f (2) = 8 + 1$

Add $8$ and $1$ .

$f (2) = 9$

The final answer is $9$ .

$9$

The $y$ value at $x = 2$ is $9$ .

$y = 9$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 2 & 17 \\ - 1 & 6 \\ \frac{1}{3} & \frac{2}{3} \\ 1 & 2 \\ 2 & 9 \end{array}$

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(\frac{1}{3}, \frac{2}{3})$

Focus: $(\frac{1}{3}, \frac{3}{4})$

Axis of Symmetry: $x = \frac{1}{3}$

Directrix: $y = \frac{7}{12}$

$\begin{array}{c} x & y \\ - 2 & 17 \\ - 1 & 6 \\ \frac{1}{3} & \frac{2}{3} \\ 1 & 2 \\ 2 & 9 \end{array}$

Step 4

Graph y=3x^2-2x+1

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