Find the x and y Intercepts f(x)=3(1/2)^x-4

$f (x) = 3 {(\frac{1}{2})}^{x} - 4$

Step 1

Find the x-intercepts.

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

$0 = 3 {(\frac{1}{2})}^{x} - 4$

Solve the equation.

Rewrite the equation as $3 {(\frac{1}{2})}^{x} - 4 = 0$ .

$3 {(\frac{1}{2})}^{x} - 4 = 0$

Simplify each term.

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

$3 \frac{1^{x}}{2^{x}} - 4 = 0$

One to any power is one.

$3 \frac{1}{2^{x}} - 4 = 0$

Combine $3$ and $\frac{1}{2^{x}}$ .

$\frac{3}{2^{x}} - 4 = 0$

Add $4$ to both sides of the equation.

$\frac{3}{2^{x}} = 4$

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

$\frac{3}{2^{x}} \cdot 2^{x} = 4 \cdot 2^{x}$

Simplify.

Simplify the left side.

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

Cancel the common factor.

$\frac{3}{2^{x}} \cdot 2^{x} = 4 \cdot 2^{x}$

Rewrite the expression.

$3 = 4 \cdot 2^{x}$

Simplify the right side.

Multiply $4 \cdot 2^{x}$ .

Rewrite $4$ as $2^{2}$ .

$3 = 2^{2} \cdot 2^{x}$

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

$3 = 2^{2 + x}$

Solve for $x$ .

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

$2^{2 + x} = 3$

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

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

Expand $\ln (2^{2 + x})$ by moving $2 + x$ outside the logarithm.

$(2 + x) \ln (2) = \ln (3)$

Simplify the left side.

Apply the distributive property.

$2 \ln (2) + x \ln (2) = \ln (3)$

Reorder $2 \ln (2)$ and $x \ln (2)$ .

$x \ln (2) + 2 \ln (2) = \ln (3)$

Move all the terms containing a logarithm to the left side of the equation.

$x \ln (2) + 2 \ln (2) - \ln (3) = 0$

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

Subtract $2 \ln (2)$ from both sides of the equation.

$x \ln (2) - \ln (3) = - 2 \ln (2)$

Add $\ln (3)$ to both sides of the equation.

$x \ln (2) = - 2 \ln (2) + \ln (3)$

Divide each term in $x \ln (2) = - 2 \ln (2) + \ln (3)$ by $\ln (2)$ and simplify.

Divide each term in $x \ln (2) = - 2 \ln (2) + \ln (3)$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{- 2 \ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Simplify the left side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{- 2 \ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Divide $x$ by $1$ .

$x = \frac{- 2 \ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Simplify the right side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$x = \frac{- 2 \ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Divide $- 2$ by $1$ .

$x = - 2 + \frac{\ln (3)}{\ln (2)}$

x-intercept(s) in point form.

x-intercept(s): $(- 2 + \frac{\ln (3)}{\ln (2)}, 0)$

Step 2

Find the y-intercepts.

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

$y = 3 {(\frac{1}{2})}^{0} - 4$

Solve the equation.

Remove parentheses.

$y = 3 {(\frac{1}{2})}^{0} - 4$

Simplify $3 {(\frac{1}{2})}^{0} - 4$ .

Simplify each term.

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

$y = 3 \frac{1^{0}}{2^{0}} - 4$

Anything raised to $0$ is $1$ .

$y = 3 \frac{1}{2^{0}} - 4$

Anything raised to $0$ is $1$ .

$y = 3 (\frac{1}{1}) - 4$

Cancel the common factor of $1$ .

Cancel the common factor.

$y = 3 (\frac{1}{1}) - 4$

Rewrite the expression.

$y = 3 \cdot 1 - 4$

Multiply $3$ by $1$ .

$y = 3 - 4$

Subtract $4$ from $3$ .

$y = - 1$

y-intercept(s) in point form.

y-intercept(s): $(0, - 1)$

Step 3

List the intersections.

x-intercept(s): $(- 2 + \frac{\ln (3)}{\ln (2)}, 0)$

y-intercept(s): $(0, - 1)$

Step 4

Find the x and y Intercepts f(x)=3(1/2)^x-4

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