Find the Points of Intersection y=-x^2+1 y=x^2

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $- x^{2} + 1 = x^{2}$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

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

$- x^{2} + 1 - x^{2} = 0$

Subtract $x^{2}$ from $- x^{2}$ .

$- 2 x^{2} + 1 = 0$

Subtract $1$ from both sides of the equation.

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

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

Divide each term in $- 2 x^{2} = - 1$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Dividing two negative values results in a positive value.

$x^{2} = \frac{1}{2}$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{1}{2}}$

Simplify $\pm \sqrt{\frac{1}{2}}$ .

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{2}}$

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

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

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

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

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

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

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

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

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

$x = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

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

$x = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

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

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$x = \pm \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$x = \pm \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{\sqrt{2}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 3

Evaluate $y$ when $x = \frac{\sqrt{2}}{2}$ .

Substitute $\frac{\sqrt{2}}{2}$ for $x$ .

$y = {(\frac{\sqrt{2}}{2})}^{2}$

Simplify ${(\frac{\sqrt{2}}{2})}^{2}$ .

Apply the product rule to $\frac{\sqrt{2}}{2}$ .

$y = \frac{{\sqrt{2}}^{2}}{2^{2}}$

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

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

$y = \frac{{(2^{\frac{1}{2}})}^{2}}{2^{2}}$

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

$y = \frac{2^{\frac{1}{2} \cdot 2}}{2^{2}}$

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

$y = \frac{2^{\frac{2}{2}}}{2^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \frac{2^{\frac{2}{2}}}{2^{2}}$

Rewrite the expression.

$y = \frac{2^{1}}{2^{2}}$

Evaluate the exponent.

$y = \frac{2}{2^{2}}$

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

$y = \frac{2}{4}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$y = \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = \frac{1}{2}$

Step 4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\frac{\sqrt{2}}{2}, \frac{1}{2})$

$(- \frac{\sqrt{2}}{2}, \frac{1}{2})$

Step 5

The result can be shown in multiple forms.

Point Form:

$(\frac{\sqrt{2}}{2}, \frac{1}{2}), (- \frac{\sqrt{2}}{2}, \frac{1}{2})$

Equation Form:

$x = \frac{\sqrt{2}}{2}, y = \frac{1}{2}$

$x = - \frac{\sqrt{2}}{2}, y = \frac{1}{2}$

Step 6

Find the Points of Intersection y=-x^2+1 y=x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.