Precalculus Examples

$y = 2^{x - 5} + 3$

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 $x$ .

$0 = 2^{x - 5} + 3$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $2^{x - 5} + 3 = 0$ .

$2^{x - 5} + 3 = 0$

Step 1.2.2

Subtract $3$ from both sides of the equation.

$2^{x - 5} = - 3$

Step 1.2.3

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

$\ln (2^{x - 5}) = \ln (- 3)$

Step 1.2.4

The equation cannot be solved because $\ln (- 3)$ is undefined.

Undefined

Step 1.2.5

There is no solution for $2^{x - 5} = - 3$

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 $y$ .

$y = 2^{(0) - 5} + 3$

Step 2.2

Simplify $2^{(0) - 5} + 3$ .

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Subtract $5$ from $0$ .

$y = 2^{- 5} + 3$

Step 2.2.1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y = \frac{1}{2^{5}} + 3$

Step 2.2.1.3

Raise $2$ to the power of $5$ .

$y = \frac{1}{32} + 3$

Step 2.2.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{32}{32}$ .

$y = \frac{1}{32} + 3 \cdot \frac{32}{32}$

Step 2.2.3

Combine $3$ and $\frac{32}{32}$ .

$y = \frac{1}{32} + \frac{3 \cdot 32}{32}$

Step 2.2.4

Combine the numerators over the common denominator.

$y = \frac{1 + 3 \cdot 32}{32}$

Step 2.2.5

Simplify the numerator.

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Step 2.2.5.1

Multiply $3$ by $32$ .

$y = \frac{1 + 96}{32}$

Step 2.2.5.2

Add $1$ and $96$ .

$y = \frac{97}{32}$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{97}{32})$

Step 3

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, \frac{97}{32})$

Step 4