Graph (y+1)^2=x-4

${(y + 1)}^{2} = x - 4$

Step 1

Rewrite the equation as $x - 4 = {(y + 1)}^{2}$ .

$x - 4 = {(y + 1)}^{2}$

Step 2

Simplify ${(y + 1)}^{2}$ .

Rewrite.

$x - 4 = 0 + 0 + {(y + 1)}^{2}$

Rewrite ${(y + 1)}^{2}$ as $(y + 1) (y + 1)$ .

$x - 4 = (y + 1) (y + 1)$

Expand $(y + 1) (y + 1)$ using the FOIL Method.

Apply the distributive property.

$x - 4 = y (y + 1) + 1 (y + 1)$

Apply the distributive property.

$x - 4 = y \cdot y + y \cdot 1 + 1 (y + 1)$

Apply the distributive property.

$x - 4 = y \cdot y + y \cdot 1 + 1 y + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$x - 4 = y^{2} + y \cdot 1 + 1 y + 1 \cdot 1$

Multiply $y$ by $1$ .

$x - 4 = y^{2} + y + 1 y + 1 \cdot 1$

Multiply $y$ by $1$ .

$x - 4 = y^{2} + y + y + 1 \cdot 1$

Multiply $1$ by $1$ .

$x - 4 = y^{2} + y + y + 1$

Add $y$ and $y$ .

$x - 4 = y^{2} + 2 y + 1$

Step 3

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$x = y^{2} + 2 y + 1 + 4$

Add $1$ and $4$ .

$x = y^{2} + 2 y + 5$

Step 4

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $y^{2} + 2 y + 5$ .

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

$a = 1$

$b = 2$

$c = 5$

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 5 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 5 - 1$

Subtract $1$ from $5$ .

$e = 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 1)}^{2} + 4$ .

${(y + 1)}^{2} + 4$

Set $x$ equal to the new right side.

$x = {(y + 1)}^{2} + 4$

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

$a = 1$

$h = 4$

$k = - 1$

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

Opens Right

Find the vertex $(h, k)$ .

$(4, - 1)$

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 $1$ .

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 x-coordinate $h$ if the parabola opens left or right.

$(h + p, k)$

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(\frac{17}{4}, - 1)$

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

$y = - 1$

Find the directrix.

The directrix of a parabola is the vertical line found by subtracting $p$ from the x-coordinate $h$ of the vertex if the parabola opens left or right.

$x = h - p$

Substitute the known values of $p$ and $h$ into the formula and simplify.

$x = \frac{15}{4}$

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

Direction: Opens Right

Vertex: $(4, - 1)$

Focus: $(\frac{17}{4}, - 1)$

Axis of Symmetry: $y = - 1$

Directrix: $x = \frac{15}{4}$

Direction: Opens Right

Vertex: $(4, - 1)$

Focus: $(\frac{17}{4}, - 1)$

Axis of Symmetry: $y = - 1$

Directrix: $x = \frac{15}{4}$

Step 5

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.

Substitute the $x$ value $5$ into $f (x) = \sqrt{x - 4} - 1$ . In this case, the point is $(5, 0)$ .

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

$f (5) = \sqrt{(5) - 4} - 1$

Simplify the result.

Simplify each term.

Subtract $4$ from $5$ .

$f (5) = \sqrt{1} - 1$

Any root of $1$ is $1$ .

$f (5) = 1 - 1$

Subtract $1$ from $1$ .

$f (5) = 0$

The final answer is $0$ .

$y = 0$

Convert $0$ to decimal.

$= 0$

Substitute the $x$ value $5$ into $f (x) = - \sqrt{x - 4} - 1$ . In this case, the point is $(5, - 2)$ .

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

$f (5) = - \sqrt{(5) - 4} - 1$

Simplify the result.

Simplify each term.

Subtract $4$ from $5$ .

$f (5) = - \sqrt{1} - 1$

Any root of $1$ is $1$ .

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

Multiply $- 1$ by $1$ .

$f (5) = - 1 - 1$

Subtract $1$ from $- 1$ .

$f (5) = - 2$

The final answer is $- 2$ .

$y = - 2$

Convert $- 2$ to decimal.

$= - 2$

Substitute the $x$ value $6$ into $f (x) = \sqrt{x - 4} - 1$ . In this case, the point is $(6, 0.41421356)$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = \sqrt{(6) - 4} - 1$

Simplify the result.

Subtract $4$ from $6$ .

$f (6) = \sqrt{2} - 1$

The final answer is $\sqrt{2} - 1$ .

$y = \sqrt{2} - 1$

Convert $\sqrt{2} - 1$ to decimal.

$= 0.41421356$

Substitute the $x$ value $6$ into $f (x) = - \sqrt{x - 4} - 1$ . In this case, the point is $(6, - 2.41421356)$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = - \sqrt{(6) - 4} - 1$

Simplify the result.

Subtract $4$ from $6$ .

$f (6) = - \sqrt{2} - 1$

The final answer is $- \sqrt{2} - 1$ .

$y = - \sqrt{2} - 1$

Convert $- \sqrt{2} - 1$ to decimal.

$= - 2.41421356$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ 4 & - 1 \\ 5 & 0 \\ 5 & - 2 \\ 6 & 0.41 \\ 6 & - 2.41 \end{array}$

Step 6

Graph the parabola using its properties and the selected points.

Direction: Opens Right

Vertex: $(4, - 1)$

Focus: $(\frac{17}{4}, - 1)$

Axis of Symmetry: $y = - 1$

Directrix: $x = \frac{15}{4}$

$\begin{array}{c} x & y \\ 4 & - 1 \\ 5 & 0 \\ 5 & - 2 \\ 6 & 0.41 \\ 6 & - 2.41 \end{array}$

Step 7

Graph (y+1)^2=x-4

Related Questions

Support LCMGCF.com by disabling your adblocker.