Graph y=x^2-6x-7

$y = x^{2} - 6 x - 7$

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $x^{2} - 6 x - 7$ .

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

$a = 1$

$b = - 6$

$c = - 7$

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

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

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3$ by $1$ .

$d = - 3$

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

$e = - 7 - \frac{{(- 6)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = - 7 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 7 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = - 7 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = - 7 - 9$

Subtract $9$ from $- 7$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 3)}^{2} - 16$ .

${(x - 3)}^{2} - 16$

Set $y$ equal to the new right side.

$y = {(x - 3)}^{2} - 16$

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

$a = 1$

$h = 3$

$k = - 16$

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

Opens Up

Find the vertex $(h, k)$ .

$(3, - 16)$

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.

$(3, - \frac{63}{4})$

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

$x = 3$

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.

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

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

Direction: Opens Up

Vertex: $(3, - 16)$

Focus: $(3, - \frac{63}{4})$

Axis of Symmetry: $x = 3$

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

Direction: Opens Up

Vertex: $(3, - 16)$

Focus: $(3, - \frac{63}{4})$

Axis of Symmetry: $x = 3$

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 $2$ in the expression.

$f (2) = {(2)}^{2} - 6 \cdot 2 - 7$

Simplify the result.

Simplify each term.

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

$f (2) = 4 - 6 \cdot 2 - 7$

Multiply $- 6$ by $2$ .

$f (2) = 4 - 12 - 7$

Simplify by subtracting numbers.

Subtract $12$ from $4$ .

$f (2) = - 8 - 7$

Subtract $7$ from $- 8$ .

$f (2) = - 15$

The final answer is $- 15$ .

$- 15$

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

$y = - 15$

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

$f (1) = {(1)}^{2} - 6 \cdot 1 - 7$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 1 - 6 \cdot 1 - 7$

Multiply $- 6$ by $1$ .

$f (1) = 1 - 6 - 7$

Simplify by subtracting numbers.

Subtract $6$ from $1$ .

$f (1) = - 5 - 7$

Subtract $7$ from $- 5$ .

$f (1) = - 12$

The final answer is $- 12$ .

$- 12$

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

$y = - 12$

Replace the variable $x$ with $4$ in the expression.

$f (4) = {(4)}^{2} - 6 \cdot 4 - 7$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$f (4) = 16 - 6 \cdot 4 - 7$

Multiply $- 6$ by $4$ .

$f (4) = 16 - 24 - 7$

Simplify by subtracting numbers.

Subtract $24$ from $16$ .

$f (4) = - 8 - 7$

Subtract $7$ from $- 8$ .

$f (4) = - 15$

The final answer is $- 15$ .

$- 15$

The $y$ value at $x = 4$ is $- 15$ .

$y = - 15$

Replace the variable $x$ with $5$ in the expression.

$f (5) = {(5)}^{2} - 6 \cdot 5 - 7$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$f (5) = 25 - 6 \cdot 5 - 7$

Multiply $- 6$ by $5$ .

$f (5) = 25 - 30 - 7$

Simplify by subtracting numbers.

Subtract $30$ from $25$ .

$f (5) = - 5 - 7$

Subtract $7$ from $- 5$ .

$f (5) = - 12$

The final answer is $- 12$ .

$- 12$

The $y$ value at $x = 5$ is $- 12$ .

$y = - 12$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ 1 & - 12 \\ 2 & - 15 \\ 3 & - 16 \\ 4 & - 15 \\ 5 & - 12 \end{array}$

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(3, - 16)$

Focus: $(3, - \frac{63}{4})$

Axis of Symmetry: $x = 3$

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

$\begin{array}{c} x & y \\ 1 & - 12 \\ 2 & - 15 \\ 3 & - 16 \\ 4 & - 15 \\ 5 & - 12 \end{array}$

Step 4

Graph y=x^2-6x-7

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