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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

Rewrite the equation as $- 2 \cdot 2^{\frac{1}{3} x} + 3 = 0$ .

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

Simplify each term.

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

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

Multiply $- 2 \cdot 2^{\frac{x}{3}}$ .

Factor out negative.

$- (2 \cdot 2^{\frac{x}{3}}) + 3 = 0$

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

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

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

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

Subtract $3$ from both sides of the equation.

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

Divide each term in $- 2^{1 + \frac{x}{3}} = - 3$ by $- 1$ and simplify.

Divide each term in $- 2^{1 + \frac{x}{3}} = - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $2^{1 + \frac{x}{3}}$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$2^{1 + \frac{x}{3}} = 3$

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

$\ln (2^{1 + \frac{x}{3}}) = \ln (3)$

Expand $\ln (2^{1 + \frac{x}{3}})$ by moving $1 + \frac{x}{3}$ outside the logarithm.

$(1 + \frac{x}{3}) \ln (2) = \ln (3)$

Simplify the left side.

Simplify $(1 + \frac{x}{3}) \ln (2)$ .

Apply the distributive property.

$1 \ln (2) + \frac{x}{3} \ln (2) = \ln (3)$

Multiply $\ln (2)$ by $1$ .

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

Combine $\frac{x}{3}$ and $\ln (2)$ .

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

Reorder $\ln (2)$ and $\frac{x \ln (2)}{3}$ .

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

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

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{x \ln (2)}{3} + \ln (\frac{2}{3}) = 0$

Subtract $\ln (\frac{2}{3})$ from both sides of the equation.

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

Multiply both sides of the equation by $3$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Multiply $- 1$ by $3$ .

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

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

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

Simplify each term.

Multiply $\frac{1}{3}$ by $0$ .

$y = - 2 \cdot 2^{0} + 3$

Anything raised to $0$ is $1$ .

$y = - 2 \cdot 1 + 3$

Multiply $- 2$ by $1$ .

$y = - 2 + 3$

Add $- 2$ and $3$ .

$y = 1$

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4

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

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