Graph y=-(-x)^2+1

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

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Simplify each term.

Apply the product rule to $- x$ .

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

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Move ${(- 1)}^{2}$ .

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

Multiply ${(- 1)}^{2}$ by $- 1$ .

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

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $1$ .

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

Raise $- 1$ to the power of $3$ .

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

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

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

$a = - 1$

$b = 0$

$c = 1$

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

$d = \frac{0}{2 \cdot - 1}$

Simplify the right side.

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

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{0}{- 1}$ .

$d = - 1 \cdot 0$

Rewrite $- 1 \cdot 0$ as $- 0$ .

$d = - 0$

Multiply $- 1$ by $0$ .

$d = 0$

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

$e = 1 - \frac{0^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = 1 - \frac{0}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 1 - \frac{0}{- 4}$

Divide $0$ by $- 4$ .

$e = 1 - 0$

Multiply $- 1$ by $0$ .

$e = 1 + 0$

Add $1$ and $0$ .

$e = 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 0)}^{2} + 1$ .

$- {(x + 0)}^{2} + 1$

Set $y$ equal to the new right side.

$y = - {(x + 0)}^{2} + 1$

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

$a = - 1$

$h = 0$

$k = 1$

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

Opens Down

Find the vertex $(h, k)$ .

$(0, 1)$

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.

$(0, \frac{3}{4})$

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.

$y = \frac{5}{4}$

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

Direction: Opens Down

Vertex: $(0, 1)$

Focus: $(0, \frac{3}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{5}{4}$

Direction: Opens Down

Vertex: $(0, 1)$

Focus: $(0, \frac{3}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{5}{4}$

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.

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

Simplify the result.

Simplify each term.

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

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

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

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

Raise $- 1$ to the power of $3$ .

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

Add $- 1$ and $1$ .

$f (- 1) = 0$

The final answer is $0$ .

$0$

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

$y = 0$

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

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

Simplify the result.

Simplify each term.

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

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

Multiply $- 1$ by $4$ .

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

Add $- 4$ and $1$ .

$f (- 2) = - 3$

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.

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

Simplify the result.

Simplify each term.

One to any power is one.

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

Multiply $- 1$ by $1$ .

$f (1) = - 1 + 1$

Add $- 1$ and $1$ .

$f (1) = 0$

The final answer is $0$ .

$0$

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

$y = 0$

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

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

Simplify the result.

Simplify each term.

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

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

Multiply $- 1$ by $4$ .

$f (2) = - 4 + 1$

Add $- 4$ and $1$ .

$f (2) = - 3$

The final answer is $- 3$ .

$- 3$

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

$y = - 3$

Graph the parabola using its properties and the selected points.

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

Graph the parabola using its properties and the selected points.

Direction: Opens Down

Vertex: $(0, 1)$

Focus: $(0, \frac{3}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = \frac{5}{4}$

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

Graph y=-(-x)^2+1

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