Graph x^2+((5y)/4- square root of |x|)^2=1

$x^{2} + {(\frac{5 y}{4} - \sqrt{| x |})}^{2} = 1$

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

Subtract $x^{2}$ from both sides of the equation.

$| x | + \frac{25 y^{2}}{16} - \frac{5 y \sqrt{| x |}}{2} = 1 - x^{2}$

Subtract $| x |$ from both sides of the equation.

$\frac{25 y^{2}}{16} - \frac{5 y \sqrt{| x |}}{2} = 1 - x^{2} - | x |$

Find the absolute value vertex. In this case, the vertex for $\frac{25 y^{2}}{16} - \frac{5 y \sqrt{| x |}}{2} = 1 - x^{2} - | x |$ is $(0, 1)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x$ equal to $0$ . In this case, $x = 0$ .

$x = 0$

Replace the variable $x$ with $0$ in the expression.

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

Simplify $1 - {(0)}^{2} - | 0 |$ .

Simplify each term.

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

$y = 1 - 0 - | 0 |$

Multiply $- 1$ by $0$ .

$y = 1 + 0 - | 0 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 1 + 0 - 0$

Multiply $- 1$ by $0$ .

$y = 1 + 0 + 0$

Simplify by adding numbers.

Add $1$ and $0$ .

$y = 1 + 0$

Add $1$ and $0$ .

$y = 1$

The absolute value vertex is $(0, 1)$ .

$(0, 1)$

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

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

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

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

Simplify the result.

Simplify each term.

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

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

Multiply $- 1$ by $4$ .

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

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

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

Multiply $- 1$ by $2$ .

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

Simplify by subtracting numbers.

Subtract $4$ from $1$ .

$f (- 2) = - 3 - 2$

Subtract $2$ from $- 3$ .

$f (- 2) = - 5$

The final answer is $- 5$ .

$y = - 5$

Substitute the $x$ value $- 1$ into $f (x) = 1 - x^{2} - | x |$ . In this case, the point is $(- 1, - 1)$ .

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

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

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 $1$ .

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

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

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

Add $1$ and $2$ .

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

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

$f (- 1) = 1 - 1 - | - 1 |$

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify by subtracting numbers.

Subtract $1$ from $1$ .

$f (- 1) = 0 - 1$

Subtract $1$ from $0$ .

$f (- 1) = - 1$

The final answer is $- 1$ .

$y = - 1$

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

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

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

Simplify the result.

Simplify each term.

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

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

Multiply $- 1$ by $4$ .

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

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Multiply $- 1$ by $2$ .

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

Simplify by subtracting numbers.

Subtract $4$ from $1$ .

$f (2) = - 3 - 2$

Subtract $2$ from $- 3$ .

$f (2) = - 5$

The final answer is $- 5$ .

$y = - 5$

The absolute value can be graphed using the points around the vertex $(0, 1), (- 2, - 5), (- 1, - 1), (1, - 1), (2, - 5)$

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

Graph x^2+((5y)/4- square root of |x|)^2=1

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