Graph f(x)=-x^2+6x-5

$f (x) = - x^{2} + 6 x - 5$

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $- x^{2} + 6 x - 5$ .

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

$a = - 1$

$b = 6$

$c = - 5$

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

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

Simplify the right side.

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

Factor $2$ out of $6$ .

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

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$d = - 3$

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

$e = - 5 - \frac{6^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = - 5 - \frac{36}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 5 - \frac{36}{- 4}$

Divide $36$ by $- 4$ .

$e = - 5 - - 9$

Multiply $- 1$ by $- 9$ .

$e = - 5 + 9$

Add $- 5$ and $9$ .

$e = 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 3)}^{2} + 4$ .

$- {(x - 3)}^{2} + 4$

Set $y$ equal to the new right side.

$y = - {(x - 3)}^{2} + 4$

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

$a = - 1$

$h = 3$

$k = 4$

Since the value of $a$ is negative, the parabola opens down.

Opens Down

Find the vertex $(h, k)$ .

$(3, 4)$

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 $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- 1 (- 1)}{4 \cdot - 1}$

Move the negative in front of the fraction.

$- \frac{1}{4}$

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.

$(3, \frac{15}{4})$

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.

$y = \frac{17}{4}$

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

Direction: Opens Down

Vertex: $(3, 4)$

Focus: $(3, \frac{15}{4})$

Axis of Symmetry: $x = 3$

Directrix: $y = \frac{17}{4}$

Direction: Opens Down

Vertex: $(3, 4)$

Focus: $(3, \frac{15}{4})$

Axis of Symmetry: $x = 3$

Directrix: $y = \frac{17}{4}$

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.

$f (2) = - {(2)}^{2} + 6 (2) - 5$

Simplify the result.

Simplify each term.

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

$f (2) = - 1 \cdot 4 + 6 (2) - 5$

Multiply $- 1$ by $4$ .

$f (2) = - 4 + 6 (2) - 5$

Multiply $6$ by $2$ .

$f (2) = - 4 + 12 - 5$

Simplify by adding and subtracting.

Add $- 4$ and $12$ .

$f (2) = 8 - 5$

Subtract $5$ from $8$ .

$f (2) = 3$

The final answer is $3$ .

$3$

The $y$ value at $x = 2$ is $3$ .

$y = 3$

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

$f (1) = - {(1)}^{2} + 6 (1) - 5$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = - 1 \cdot 1 + 6 (1) - 5$

Multiply $- 1$ by $1$ .

$f (1) = - 1 + 6 (1) - 5$

Multiply $6$ by $1$ .

$f (1) = - 1 + 6 - 5$

Simplify by adding and subtracting.

Add $- 1$ and $6$ .

$f (1) = 5 - 5$

Subtract $5$ from $5$ .

$f (1) = 0$

The final answer is $0$ .

$0$

The $y$ value at $x = 1$ is $0$ .

$y = 0$

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

$f (4) = - {(4)}^{2} + 6 (4) - 5$

Simplify the result.

Simplify each term.

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

$f (4) = - 1 \cdot 16 + 6 (4) - 5$

Multiply $- 1$ by $16$ .

$f (4) = - 16 + 6 (4) - 5$

Multiply $6$ by $4$ .

$f (4) = - 16 + 24 - 5$

Simplify by adding and subtracting.

Add $- 16$ and $24$ .

$f (4) = 8 - 5$

Subtract $5$ from $8$ .

$f (4) = 3$

The final answer is $3$ .

$3$

The $y$ value at $x = 4$ is $3$ .

$y = 3$

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

$f (5) = - {(5)}^{2} + 6 (5) - 5$

Simplify the result.

Simplify each term.

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

$f (5) = - 1 \cdot 25 + 6 (5) - 5$

Multiply $- 1$ by $25$ .

$f (5) = - 25 + 6 (5) - 5$

Multiply $6$ by $5$ .

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

Simplify by adding and subtracting.

Add $- 25$ and $30$ .

$f (5) = 5 - 5$

Subtract $5$ from $5$ .

$f (5) = 0$

The final answer is $0$ .

$0$

The $y$ value at $x = 5$ is $0$ .

$y = 0$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 3 \\ 3 & 4 \\ 4 & 3 \\ 5 & 0 \end{array}$

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Down

Vertex: $(3, 4)$

Focus: $(3, \frac{15}{4})$

Axis of Symmetry: $x = 3$

Directrix: $y = \frac{17}{4}$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 3 \\ 3 & 4 \\ 4 & 3 \\ 5 & 0 \end{array}$

Step 4

Graph f(x)=-x^2+6x-5

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