Graph y=x^2-2x-15

$y = x^{2} - 2 x - 15$

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $x^{2} - 2 x - 15$ .

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

$a = 1$

$b = - 2$

$c = - 15$

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

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

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

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

$e = - 15 - \frac{{(- 2)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = - 15 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 15 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = - 15 - \frac{4}{4}$

Rewrite the expression.

$e = - 15 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = - 15 - 1$

Subtract $1$ from $- 15$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 1)}^{2} - 16$ .

${(x - 1)}^{2} - 16$

Set $y$ equal to the new right side.

$y = {(x - 1)}^{2} - 16$

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

$a = 1$

$h = 1$

$k = - 16$

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

Opens Up

Find the vertex $(h, k)$ .

$(1, - 16)$

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.

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

Cancel the common factor of $1$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{4}$

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.

$(1, - \frac{63}{4})$

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

$x = 1$

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.

$y = - \frac{65}{4}$

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

Direction: Opens Up

Vertex: $(1, - 16)$

Focus: $(1, - \frac{63}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{65}{4}$

Direction: Opens Up

Vertex: $(1, - 16)$

Focus: $(1, - \frac{63}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{65}{4}$

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 $0$ in the expression.

$f (0) = {(0)}^{2} - 2 \cdot 0 - 15$

Simplify the result.

Simplify each term.

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

$f (0) = 0 - 2 \cdot 0 - 15$

Multiply $- 2$ by $0$ .

$f (0) = 0 + 0 - 15$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$f (0) = 0 - 15$

Subtract $15$ from $0$ .

$f (0) = - 15$

The final answer is $- 15$ .

$- 15$

The $y$ value at $x = 0$ is $- 15$ .

$y = - 15$

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

$f (- 1) = {(- 1)}^{2} - 2 \cdot - 1 - 15$

Simplify the result.

Simplify each term.

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

$f (- 1) = 1 - 2 \cdot - 1 - 15$

Multiply $- 2$ by $- 1$ .

$f (- 1) = 1 + 2 - 15$

Simplify by adding and subtracting.

Add $1$ and $2$ .

$f (- 1) = 3 - 15$

Subtract $15$ from $3$ .

$f (- 1) = - 12$

The final answer is $- 12$ .

$- 12$

The $y$ value at $x = - 1$ is $- 12$ .

$y = - 12$

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

$f (2) = {(2)}^{2} - 2 \cdot 2 - 15$

Simplify the result.

Simplify each term.

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

$f (2) = 4 - 2 \cdot 2 - 15$

Multiply $- 2$ by $2$ .

$f (2) = 4 - 4 - 15$

Simplify by subtracting numbers.

Subtract $4$ from $4$ .

$f (2) = 0 - 15$

Subtract $15$ from $0$ .

$f (2) = - 15$

The final answer is $- 15$ .

$- 15$

The $y$ value at $x = 2$ is $- 15$ .

$y = - 15$

Replace the variable $x$ with $3$ in the expression.

$f (3) = {(3)}^{2} - 2 \cdot 3 - 15$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$f (3) = 9 - 2 \cdot 3 - 15$

Multiply $- 2$ by $3$ .

$f (3) = 9 - 6 - 15$

Simplify by subtracting numbers.

Subtract $6$ from $9$ .

$f (3) = 3 - 15$

Subtract $15$ from $3$ .

$f (3) = - 12$

The final answer is $- 12$ .

$- 12$

The $y$ value at $x = 3$ is $- 12$ .

$y = - 12$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 1 & - 12 \\ 0 & - 15 \\ 1 & - 16 \\ 2 & - 15 \\ 3 & - 12 \end{array}$

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(1, - 16)$

Focus: $(1, - \frac{63}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{65}{4}$

$\begin{array}{c} x & y \\ - 1 & - 12 \\ 0 & - 15 \\ 1 & - 16 \\ 2 & - 15 \\ 3 & - 12 \end{array}$

Step 4

Graph y=x^2-2x-15

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