Algebra Examples

f (x) = 5 x^{2} - 5 x + 1

$f (x) = 5 x^{2} - 5 x + 1$

Step 1

The minimum of a quadratic function occurs at

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$ . If

a

$a$ is positive, the minimum value of the function is

f (- \frac{b}{2 a})

$f (- \frac{b}{2 a})$ .

f_{min}

$f_{min}$

x = a x^{2} + b x + c

$x = a x^{2} + b x + c$ occurs at

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$

Step 2

Find the value of

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$ .

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Step 2.1

Substitute in the values of

a

$a$ and

b

$b$ .

x = - \frac{- 5}{2 (5)}

$x = - \frac{- 5}{2 (5)}$

Step 2.2

Remove parentheses.

x = - \frac{- 5}{2 (5)}

$x = - \frac{- 5}{2 (5)}$

Step 2.3

Simplify

- \frac{- 5}{2 (5)}

$- \frac{- 5}{2 (5)}$ .

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Step 2.3.1

Cancel the common factor of

- 5

$- 5$ and

5

$5$ .

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Step 2.3.1.1

Factor

5

$5$ out of

- 5

$- 5$ .

x = - \frac{5 \cdot - 1}{2 \cdot 5}

$x = - \frac{5 \cdot - 1}{2 \cdot 5}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor

5

$5$ out of

2 \cdot 5

$2 \cdot 5$ .

x = - \frac{5 \cdot - 1}{5 \cdot 2}

$x = - \frac{5 \cdot - 1}{5 \cdot 2}$

Step 2.3.1.2.2

Cancel the common factor.

$x = - \frac{5 \cdot - 1}{5 \cdot 2}$

Step 2.3.1.2.3

Rewrite the expression.

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

Step 2.3.2

Move the negative in front of the fraction.

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

Step 2.3.3

Multiply

$- - \frac{1}{2}$ .

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Step 2.3.3.1

Multiply

$- 1$ by

$- 1$ .

$x = 1 (\frac{1}{2})$

Step 2.3.3.2

Multiply

$\frac{1}{2}$ by

$1$ .

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

Step 3

Evaluate

$f (\frac{1}{2})$ .

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Step 3.1

Replace the variable

$x$ with

$\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = 5 {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 1$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to

$\frac{1}{2}$ .

$f (\frac{1}{2}) = 5 (\frac{1^{2}}{2^{2}}) - 5 (\frac{1}{2}) + 1$

Step 3.2.1.2

One to any power is one.

$f (\frac{1}{2}) = 5 (\frac{1}{2^{2}}) - 5 (\frac{1}{2}) + 1$

Step 3.2.1.3

Raise

$2$ to the power of

$2$ .

$f (\frac{1}{2}) = 5 (\frac{1}{4}) - 5 (\frac{1}{2}) + 1$

Step 3.2.1.4

Combine

$5$ and

$\frac{1}{4}$ .

$f (\frac{1}{2}) = \frac{5}{4} - 5 (\frac{1}{2}) + 1$

Step 3.2.1.5

Combine

$- 5$ and

$\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} + \frac{- 5}{2} + 1$

Step 3.2.1.6

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5}{2} + 1$

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Multiply

$\frac{5}{2}$ by

$\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} - (\frac{5}{2} \cdot \frac{2}{2}) + 1$

Step 3.2.2.2

Multiply

$\frac{5}{2}$ by

$\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + 1$

Step 3.2.2.3

Write

$1$ as a fraction with denominator

$1$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{1}{1}$

Step 3.2.2.4

Multiply

$\frac{1}{1}$ by

$\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{1}{1} \cdot \frac{4}{4}$

Step 3.2.2.5

Multiply

$\frac{1}{1}$ by

$\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{4}{4}$

Step 3.2.2.6

Multiply

$2$ by

$2$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{4} + \frac{4}{4}$

Step 3.2.3

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{5 - 5 \cdot 2 + 4}{4}$

Step 3.2.4

Simplify the expression.

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Step 3.2.4.1

Multiply

$- 5$ by

$2$ .

$f (\frac{1}{2}) = \frac{5 - 10 + 4}{4}$

Step 3.2.4.2

Subtract

$10$ from

$5$ .

$f (\frac{1}{2}) = \frac{- 5 + 4}{4}$

Step 3.2.4.3

Add

$- 5$ and

$4$ .

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

Step 3.2.4.4

Move the negative in front of the fraction.

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

Step 3.2.5

The final answer is

$- \frac{1}{4}$ .

$- \frac{1}{4}$

Step 4

Use the

$x$ and

$y$ values to find where the minimum occurs.

$(\frac{1}{2}, - \frac{1}{4})$

Step 5

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