Graph x^2=-16y

$x^{2} = - 16 y$

Step 1

Solve for $y$ .

Rewrite the equation as $- 16 y = x^{2}$ .

$- 16 y = x^{2}$

Divide each term in $- 16 y = x^{2}$ by $- 16$ and simplify.

Divide each term in $- 16 y = x^{2}$ by $- 16$ .

$\frac{- 16 y}{- 16} = \frac{x^{2}}{- 16}$

Simplify the left side.

Cancel the common factor of $- 16$ .

Cancel the common factor.

$\frac{- 16 y}{- 16} = \frac{x^{2}}{- 16}$

Divide $y$ by $1$ .

$y = \frac{x^{2}}{- 16}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{x^{2}}{16}$

Step 2

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $- \frac{x^{2}}{16}$ .

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

$a = - \frac{1}{16}$

$b = 0$

$c = 0$

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

$d = \frac{0}{2 (- \frac{1}{16})}$

Simplify the right side.

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

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 (- \frac{1}{16})}$

Cancel the common factors.

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 (- \frac{1}{16})}$

Rewrite the expression.

$d = \frac{0}{- \frac{1}{16}}$

Multiply the numerator by the reciprocal of the denominator.

$d = 0 (- 1 \cdot 16)$

Multiply $0 (- 1 \cdot 16)$ .

Multiply $- 1$ by $16$ .

$d = 0 \cdot - 16$

Multiply $0$ by $- 16$ .

$d = 0$

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

$e = 0 - \frac{0^{2}}{4 (- \frac{1}{16})}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{0}{4 (- \frac{1}{16})}$

Simplify the denominator.

Multiply $- 1$ by $4$ .

$e = 0 - \frac{0}{- 4 (\frac{1}{16})}$

Combine $- 4$ and $\frac{1}{16}$ .

$e = 0 - \frac{0}{\frac{- 4}{16}}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $- 4$ and $16$ .

Factor $4$ out of $- 4$ .

$e = 0 - \frac{0}{\frac{4 (- 1)}{16}}$

Cancel the common factors.

Factor $4$ out of $16$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

$e = 0 - (0 (- 1 \cdot 4))$

Multiply $- (0 (- 1 \cdot 4))$ .

Multiply $- 1$ by $4$ .

$e = 0 - (0 \cdot - 4)$

Multiply $0$ by $- 4$ .

$e = 0 - 0$

Multiply $- 1$ by $0$ .

$e = 0 + 0$

Add $0$ and $0$ .

$e = 0$

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

$- \frac{1}{16} x^{2}$

Set $y$ equal to the new right side.

$y = - \frac{1}{16} x^{2}$

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

$a = - \frac{1}{16}$

$h = 0$

$k = 0$

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

Opens Down

Find the vertex $(h, k)$ .

$(0, 0)$

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.

$\frac{1}{4 \cdot (- \frac{1}{16})}$

Simplify.

Cancel the common factor of $1$ and $- 1$ .

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

$\frac{- 1 (- 1)}{4 \cdot (- \frac{1}{16})}$

Move the negative in front of the fraction.

$- \frac{1}{4 (\frac{1}{16})}$

Combine $4$ and $\frac{1}{16}$ .

$- \frac{1}{\frac{4}{16}}$

Cancel the common factor of $4$ and $16$ .

Factor $4$ out of $4$ .

$- \frac{1}{\frac{4 (1)}{16}}$

Cancel the common factors.

Factor $4$ out of $16$ .

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

Cancel the common factor.

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

Rewrite the expression.

$- \frac{1}{\frac{1}{4}}$

Multiply the numerator by the reciprocal of the denominator.

$- (1 \cdot 4)$

Multiply $- (1 \cdot 4)$ .

Multiply $4$ by $1$ .

$- 1 \cdot 4$

Multiply $- 1$ by $4$ .

$- 4$

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.

$(0, - 4)$

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.

$y = 4$

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

Direction: Opens Down

Vertex: $(0, 0)$

Focus: $(0, - 4)$

Axis of Symmetry: $x = 0$

Directrix: $y = 4$

Direction: Opens Down

Vertex: $(0, 0)$

Focus: $(0, - 4)$

Axis of Symmetry: $x = 0$

Directrix: $y = 4$

Step 3

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.

$f (- 1) = - \frac{{(- 1)}^{2}}{16}$

Simplify the result.

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

$f (- 1) = - \frac{1}{16}$

The final answer is $- \frac{1}{16}$ .

$- \frac{1}{16}$

The $y$ value at $x = - 1$ is $- \frac{1}{16}$ .

$y = - \frac{1}{16}$

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

$f (- 2) = - \frac{{(- 2)}^{2}}{16}$

Simplify the result.

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

$f (- 2) = - \frac{4}{16}$

Cancel the common factor of $4$ and $16$ .

Factor $4$ out of $4$ .

$f (- 2) = - \frac{4 (1)}{16}$

Cancel the common factors.

Factor $4$ out of $16$ .

$f (- 2) = - \frac{4 \cdot 1}{4 \cdot 4}$

Cancel the common factor.

$f (- 2) = - \frac{4 \cdot 1}{4 \cdot 4}$

Rewrite the expression.

$f (- 2) = - \frac{1}{4}$

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

The $y$ value at $x = - 2$ is $- \frac{1}{4}$ .

$y = - \frac{1}{4}$

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

$f (1) = - \frac{{(1)}^{2}}{16}$

Simplify the result.

One to any power is one.

$f (1) = - \frac{1}{16}$

The final answer is $- \frac{1}{16}$ .

$- \frac{1}{16}$

The $y$ value at $x = 1$ is $- \frac{1}{16}$ .

$y = - \frac{1}{16}$

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

$f (2) = - \frac{{(2)}^{2}}{16}$

Simplify the result.

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

$f (2) = - \frac{4}{16}$

Cancel the common factor of $4$ and $16$ .

Factor $4$ out of $4$ .

$f (2) = - \frac{4 (1)}{16}$

Cancel the common factors.

Factor $4$ out of $16$ .

$f (2) = - \frac{4 \cdot 1}{4 \cdot 4}$

Cancel the common factor.

$f (2) = - \frac{4 \cdot 1}{4 \cdot 4}$

Rewrite the expression.

$f (2) = - \frac{1}{4}$

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

The $y$ value at $x = 2$ is $- \frac{1}{4}$ .

$y = - \frac{1}{4}$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 2 & - \frac{1}{4} \\ - 1 & - \frac{1}{16} \\ 0 & 0 \\ 1 & - \frac{1}{16} \\ 2 & - \frac{1}{4} \end{array}$

Step 4

Graph the parabola using its properties and the selected points.

Direction: Opens Down

Vertex: $(0, 0)$

Focus: $(0, - 4)$

Axis of Symmetry: $x = 0$

Directrix: $y = 4$

$\begin{array}{c} x & y \\ - 2 & - \frac{1}{4} \\ - 1 & - \frac{1}{16} \\ 0 & 0 \\ 1 & - \frac{1}{16} \\ 2 & - \frac{1}{4} \end{array}$

Step 5

Graph x^2=-16y

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