Algebra Examples

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

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = {(\frac{1}{2})}^{x} + 2$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as ${(\frac{1}{2})}^{x} + 2 = 0$ .

${(\frac{1}{2})}^{x} + 2 = 0$

Step 1.2.2

Simplify each term.

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Step 1.2.2.1

Apply the product rule to $\frac{1}{2}$ .

$\frac{1^{x}}{2^{x}} + 2 = 0$

Step 1.2.2.2

One to any power is one.

$\frac{1}{2^{x}} + 2 = 0$

Step 1.2.3

Subtract $2$ from both sides of the equation.

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

Step 1.2.4

Multiply both sides by $2^{x}$ .

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

Step 1.2.5

Simplify.

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Step 1.2.5.1

Simplify the left side.

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Step 1.2.5.1.1

Cancel the common factor of $2^{x}$ .

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Step 1.2.5.1.1.1

Cancel the common factor.

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

Step 1.2.5.1.1.2

Rewrite the expression.

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

Step 1.2.5.2

Simplify the right side.

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Step 1.2.5.2.1

Multiply $- 2 \cdot 2^{x}$ .

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Step 1.2.5.2.1.1

Factor out negative.

$1 = - (2 \cdot 2^{x})$

Step 1.2.5.2.1.2

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

$1 = - (2^{1} \cdot 2^{x})$

Step 1.2.5.2.1.3

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

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

Step 1.2.6

Solve for $x$ .

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Step 1.2.6.1

Rewrite the equation as $- 2^{1 + x} = 1$ .

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

Step 1.2.6.2

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

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Step 1.2.6.2.1

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

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

Step 1.2.6.2.2

Simplify the left side.

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Step 1.2.6.2.2.1

Dividing two negative values results in a positive value.

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

Step 1.2.6.2.2.2

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

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

Step 1.2.6.2.3

Simplify the right side.

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Step 1.2.6.2.3.1

Divide $1$ by $- 1$ .

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

Step 1.2.6.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{1 + x}) = \ln (- 1)$

Step 1.2.6.4

The equation cannot be solved because $\ln (- 1)$ is undefined.

Undefined

Step 1.2.6.5

There is no solution for $2^{1 + x} = - 1$

No solution

Step 1.3

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

x-intercept(s): $None$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(\frac{1}{2})}^{0} + 2$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = {(\frac{1}{2})}^{0} + 2$

Step 2.2.2

Simplify ${(\frac{1}{2})}^{0} + 2$ .

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Apply the product rule to $\frac{1}{2}$ .

$y = \frac{1^{0}}{2^{0}} + 2$

Step 2.2.2.1.2

Anything raised to $0$ is $1$ .

$y = \frac{1}{2^{0}} + 2$

Step 2.2.2.1.3

Anything raised to $0$ is $1$ .

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

Step 2.2.2.1.4

Divide $1$ by $1$ .

$y = 1 + 2$

Step 2.2.2.2

Add $1$ and $2$ .

$y = 3$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, 3)$

Step 3

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, 3)$

Step 4