Graph 5x^2+2y^2<=10 y>=x^2-2x+1

$5 x^{2} + 2 y^{2} \leq 10$ $y \geq x^{2} - 2 x + 1$

Solve for $y$ .

Subtract $5 x^{2}$ from both sides of the inequality.

$2 y^{2} \leq 10 - 5 x^{2}$

Divide each term in $2 y^{2} \leq 10 - 5 x^{2}$ by $2$ and simplify.

Divide each term in $2 y^{2} \leq 10 - 5 x^{2}$ by $2$ .

$\frac{2 y^{2}}{2} \leq \frac{10}{2} + \frac{- 5 x^{2}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} \leq \frac{10}{2} + \frac{- 5 x^{2}}{2}$

Divide $y^{2}$ by $1$ .

$y^{2} \leq \frac{10}{2} + \frac{- 5 x^{2}}{2}$

Simplify the right side.

Simplify each term.

Divide $10$ by $2$ .

$y^{2} \leq 5 + \frac{- 5 x^{2}}{2}$

Move the negative in front of the fraction.

$y^{2} \leq 5 - \frac{5 x^{2}}{2}$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{y^{2}} \leq \sqrt{5 - \frac{5 x^{2}}{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{5 - \frac{5 x^{2}}{2}}$

Simplify the right side.

Simplify $\sqrt{5 - \frac{5 x^{2}}{2}}$ .

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$| y | \leq \sqrt{5 \cdot \frac{2}{2} - \frac{5 x^{2}}{2}}$

Combine $5$ and $\frac{2}{2}$ .

$| y | \leq \sqrt{\frac{5 \cdot 2}{2} - \frac{5 x^{2}}{2}}$

Combine the numerators over the common denominator.

$| y | \leq \sqrt{\frac{5 \cdot 2 - 5 x^{2}}{2}}$

Factor $5$ out of $5 \cdot 2 - 5 x^{2}$ .

Factor $5$ out of $5 \cdot 2$ .

$| y | \leq \sqrt{\frac{5 (2) - 5 x^{2}}{2}}$

Factor $5$ out of $- 5 x^{2}$ .

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

Factor $5$ out of $5 (2) + 5 (- x^{2})$ .

$| y | \leq \sqrt{\frac{5 (2 - x^{2})}{2}}$

Rewrite $\sqrt{\frac{5 (2 - x^{2})}{2}}$ as $\frac{\sqrt{5 (2 - x^{2})}}{\sqrt{2}}$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})}}{\sqrt{2}}$

Multiply $\frac{\sqrt{5 (2 - x^{2})}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{5 (2 - x^{2})}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

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

Evaluate the exponent.

$| y | \leq \frac{\sqrt{5 (2 - x^{2})} \sqrt{2}}{2}$

Simplify the numerator.

Combine using the product rule for radicals.

$| y | \leq \frac{\sqrt{5 (2 - x^{2}) \cdot 2}}{2}$

Multiply $2$ by $5$ .

$| y | \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$

Write $| y | \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

$y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$

Find the domain of $y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ .

Set the radicand in $\sqrt{10 (2 - x^{2})}$ greater than or equal to $0$ to find where the expression is defined.

$10 (2 - x^{2}) \geq 0$

Solve for $y$ .

Divide each term in $10 (2 - x^{2}) \geq 0$ by $10$ and simplify.

Divide each term in $10 (2 - x^{2}) \geq 0$ by $10$ .

$\frac{10 (2 - x^{2})}{10} \geq \frac{0}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 (2 - x^{2})}{10} \geq \frac{0}{10}$

Divide $2 - x^{2}$ by $1$ .

$2 - x^{2} \geq \frac{0}{10}$

Simplify the right side.

Divide $0$ by $10$ .

$2 - x^{2} \geq 0$

Subtract $2$ from both sides of the inequality.

$- x^{2} \geq - 2$

Divide each term in $- x^{2} \geq - 2$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x^{2} \leq 2$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \leq \sqrt{2}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{2}$

Write $| x | \leq \sqrt{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq \sqrt{2}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq \sqrt{2}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{2} & x \geq 0 \\ - x \leq \sqrt{2} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{2}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{2}$

Solve $- x \leq \sqrt{2}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{2}$

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$x \geq - \sqrt{2}$

Find the intersection of $x \geq - \sqrt{2}$ and $x < 0$ .

$- \sqrt{2} \leq x < 0$

Find the union of the solutions.

$- \sqrt{2} \leq x \leq \sqrt{2}$

The domain is all values of $y$ that make the expression defined.

$[- \sqrt{2}, \sqrt{2}]$

Find the intersection of $y \geq 0$ and $[- \sqrt{2}, \sqrt{2}]$ .

$0 \leq y \leq \sqrt{2}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$

Find the domain of $- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ .

Set the radicand in $\sqrt{10 (2 - x^{2})}$ greater than or equal to $0$ to find where the expression is defined.

$10 (2 - x^{2}) \geq 0$

Solve for $y$ .

Divide each term in $10 (2 - x^{2}) \geq 0$ by $10$ and simplify.

Divide each term in $10 (2 - x^{2}) \geq 0$ by $10$ .

$\frac{10 (2 - x^{2})}{10} \geq \frac{0}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 (2 - x^{2})}{10} \geq \frac{0}{10}$

Divide $2 - x^{2}$ by $1$ .

$2 - x^{2} \geq \frac{0}{10}$

Simplify the right side.

Divide $0$ by $10$ .

$2 - x^{2} \geq 0$

Subtract $2$ from both sides of the inequality.

$- x^{2} \geq - 2$

Divide each term in $- x^{2} \geq - 2$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x^{2} \leq 2$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \leq \sqrt{2}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{2}$

Write $| x | \leq \sqrt{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq \sqrt{2}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq \sqrt{2}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{2} & x \geq 0 \\ - x \leq \sqrt{2} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{2}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{2}$

Solve $- x \leq \sqrt{2}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{2}$

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$x \geq - \sqrt{2}$

Find the intersection of $x \geq - \sqrt{2}$ and $x < 0$ .

$- \sqrt{2} \leq x < 0$

Find the union of the solutions.

$- \sqrt{2} \leq x \leq \sqrt{2}$

The domain is all values of $y$ that make the expression defined.

$[- \sqrt{2}, \sqrt{2}]$

Find the intersection of $y < 0$ and $[- \sqrt{2}, \sqrt{2}]$ .

$- \sqrt{2} \leq y < 0$

Write as a piecewise.

${\begin{cases} y \leq \frac{\sqrt{10 (2 - x^{2})}}{2} & 0 \leq y \leq \sqrt{2} \\ - y \leq \frac{\sqrt{10 (2 - x^{2})}}{2} & - \sqrt{2} \leq y < 0 \end{cases}$

Find the intersection of $y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ and $0 \leq y \leq \sqrt{2}$ .

$0 \leq y \leq N o (Min i m u m)$

Solve $- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ when $- \sqrt{2} \leq y < 0$ .

Divide each term in $- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ by $- 1$ and simplify.

Divide each term in $- y \leq \frac{\sqrt{10 (2 - x^{2})}}{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\frac{\sqrt{10 (2 - x^{2})}}{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\frac{\sqrt{10 (2 - x^{2})}}{2}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\frac{\sqrt{10 (2 - x^{2})}}{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{\sqrt{10 (2 - x^{2})}}{2}}{- 1}$ .

$y \geq - 1 \cdot \frac{\sqrt{10 (2 - x^{2})}}{2}$

Rewrite $- 1 \cdot \frac{\sqrt{10 (2 - x^{2})}}{2}$ as $- \frac{\sqrt{10 (2 - x^{2})}}{2}$ .

$y \geq - \frac{\sqrt{10 (2 - x^{2})}}{2}$

Find the intersection of $y \geq - \frac{\sqrt{10 (2 - x^{2})}}{2}$ and $- \sqrt{2} \leq y < 0$ .

No solution

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Graph $y \geq x^{2} - 2 x + 1$ .

The equation is not linear, so a constant slope does not exist.

Not Linear

Graph a solid line, then shade the area above the boundary line since $y$ is greater than $x^{2} - 2 x + 1$ .

$y \geq x^{2} - 2 x + 1$

Plot each graph on the same coordinate system.

$5 x^{2} + 2 y^{2} \leq 10$

$y \geq x^{2} - 2 x + 1$

Graph 5x^2+2y^2<=10 y>=x^2-2x+1

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