Graph y=3x^2+5

$y = 3 x^{2} + 5$

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $3 x^{2} + 5$ .

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

$a = 3$

$b = 0$

$c = 5$

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

$d = \frac{0}{2 \cdot 3}$

Simplify the right side.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot 3}$

Cancel the common factors.

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

$d = \frac{2 (0)}{2 (3)}$

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 \cdot 3}$

Rewrite the expression.

$d = \frac{0}{3}$

Cancel the common factor of $0$ and $3$ .

Factor $3$ out of $0$ .

$d = \frac{3 (0)}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$d = \frac{3 \cdot 0}{3 \cdot 1}$

Cancel the common factor.

$d = \frac{3 \cdot 0}{3 \cdot 1}$

Rewrite the expression.

$d = \frac{0}{1}$

Divide $0$ by $1$ .

$d = 0$

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

$e = 5 - \frac{0^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$e = 5 - \frac{0}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 5 - \frac{0}{12}$

Divide $0$ by $12$ .

$e = 5 - 0$

Multiply $- 1$ by $0$ .

$e = 5 + 0$

Add $5$ and $0$ .

$e = 5$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x + 0)}^{2} + 5$ .

$3 {(x + 0)}^{2} + 5$

Set $y$ equal to the new right side.

$y = 3 {(x + 0)}^{2} + 5$

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

$a = 3$

$h = 0$

$k = 5$

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

Opens Up

Find the vertex $(h, k)$ .

$(0, 5)$

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.

$(0, \frac{61}{12})$

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

$x = 0$

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.

$y = \frac{59}{12}$

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

Direction: Opens Up

Vertex: $(0, 5)$

Focus: $(0, \frac{61}{12})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{59}{12}$

Direction: Opens Up

Vertex: $(0, 5)$

Focus: $(0, \frac{61}{12})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{59}{12}$

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.

$f (- 1) = 3 {(- 1)}^{2} + 5$

Simplify the result.

Simplify each term.

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

$f (- 1) = 3 \cdot 1 + 5$

Multiply $3$ by $1$ .

$f (- 1) = 3 + 5$

Add $3$ and $5$ .

$f (- 1) = 8$

The final answer is $8$ .

$8$

The $y$ value at $x = - 1$ is $8$ .

$y = 8$

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

$f (- 2) = 3 {(- 2)}^{2} + 5$

Simplify the result.

Simplify each term.

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

$f (- 2) = 3 \cdot 4 + 5$

Multiply $3$ by $4$ .

$f (- 2) = 12 + 5$

Add $12$ and $5$ .

$f (- 2) = 17$

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.

$f (1) = 3 {(1)}^{2} + 5$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 3 \cdot 1 + 5$

Multiply $3$ by $1$ .

$f (1) = 3 + 5$

Add $3$ and $5$ .

$f (1) = 8$

The final answer is $8$ .

$8$

The $y$ value at $x = 1$ is $8$ .

$y = 8$

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

$f (2) = 3 {(2)}^{2} + 5$

Simplify the result.

Simplify each term.

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

$f (2) = 3 \cdot 4 + 5$

Multiply $3$ by $4$ .

$f (2) = 12 + 5$

Add $12$ and $5$ .

$f (2) = 17$

The final answer is $17$ .

$17$

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

$y = 17$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 2 & 17 \\ - 1 & 8 \\ 0 & 5 \\ 1 & 8 \\ 2 & 17 \end{array}$

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(0, 5)$

Focus: $(0, \frac{61}{12})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{59}{12}$

$\begin{array}{c} x & y \\ - 2 & 17 \\ - 1 & 8 \\ 0 & 5 \\ 1 & 8 \\ 2 & 17 \end{array}$

Step 4

Graph y=3x^2+5

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