Precalculus Examples

$x = 2 t - 1$ , $y = t^{3}$

Step 1

Set up the parametric equation for $x (t)$ to solve the equation for $t$ .

$x = 2 t - 1$

Step 2

Rewrite the equation as $2 t - 1 = x$ .

$2 t - 1 = x$

Step 3

Add $1$ to both sides of the equation.

$2 t = x + 1$

Step 4

Divide each term in $2 t = x + 1$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 t = x + 1$ by $2$ .

$\frac{2 t}{2} = \frac{x}{2} + \frac{1}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 t}{2} = \frac{x}{2} + \frac{1}{2}$

Step 4.2.1.2

Divide $t$ by $1$ .

$t = \frac{x}{2} + \frac{1}{2}$

Step 5

Replace $t$ in the equation for $y$ to get the equation in terms of $x$ .

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

Step 6

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

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Step 6.1

Use the Binomial Theorem.

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

Step 6.2

Simplify each term.

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Step 6.2.1

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

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

Step 6.2.2

Raise $2$ to the power of $3$ .

$y = \frac{x^{3}}{8} + 3 {(\frac{x}{2})}^{2} \frac{1}{2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.3

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

$y = \frac{x^{3}}{8} + 3 \frac{x^{2}}{2^{2}} \frac{1}{2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.4

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

$y = \frac{x^{3}}{8} + 3 \frac{x^{2}}{4} \frac{1}{2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.5

Combine $3$ and $\frac{x^{2}}{4}$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{4} \cdot \frac{1}{2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.6

Combine.

$y = \frac{x^{3}}{8} + \frac{3 x^{2} \cdot 1}{4 \cdot 2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.7

Multiply $3$ by $1$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{4 \cdot 2} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.8

Multiply $4$ by $2$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + 3 \frac{x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.9

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

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{2} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3}$

Step 6.2.10

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

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{2} \cdot \frac{1^{2}}{2^{2}} + {(\frac{1}{2})}^{3}$

Step 6.2.11

Combine.

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x \cdot 1^{2}}{2 \cdot 2^{2}} + {(\frac{1}{2})}^{3}$

Step 6.2.12

Multiply $2$ by $2^{2}$ by adding the exponents.

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Step 6.2.12.1

Multiply $2$ by $2^{2}$ .

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Step 6.2.12.1.1

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

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x \cdot 1^{2}}{2^{1} \cdot 2^{2}} + {(\frac{1}{2})}^{3}$

Step 6.2.12.1.2

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

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x \cdot 1^{2}}{2^{1 + 2}} + {(\frac{1}{2})}^{3}$

Step 6.2.12.2

Add $1$ and $2$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x \cdot 1^{2}}{2^{3}} + {(\frac{1}{2})}^{3}$

Step 6.2.13

Simplify the numerator.

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Step 6.2.13.1

One to any power is one.

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x \cdot 1}{2^{3}} + {(\frac{1}{2})}^{3}$

Step 6.2.13.2

Multiply $3$ by $1$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{2^{3}} + {(\frac{1}{2})}^{3}$

Step 6.2.14

Raise $2$ to the power of $3$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{8} + {(\frac{1}{2})}^{3}$

Step 6.2.15

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

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{8} + \frac{1^{3}}{2^{3}}$

Step 6.2.16

One to any power is one.

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{8} + \frac{1}{2^{3}}$

Step 6.2.17

Raise $2$ to the power of $3$ .

$y = \frac{x^{3}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{8} + \frac{1}{8}$