Find the Maximum/Minimum Value y=5x^2-2x+2

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

Step 1

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $- 2$ and $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{1}{5}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{1}{5})$

Multiply $\frac{1}{5}$ by $1$ .

$x = \frac{1}{5}$

Step 3

Evaluate $f (\frac{1}{5})$ .

Replace the variable $x$ with $\frac{1}{5}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{5}$ .

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

One to any power is one.

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

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

$f (\frac{1}{5}) = 5 (\frac{1}{25}) - 2 (\frac{1}{5}) + 2$

Cancel the common factor of $5$ .

Factor $5$ out of $25$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $- 2$ and $\frac{1}{5}$ .

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

Move the negative in front of the fraction.

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

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

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

Combine $2$ and $\frac{5}{5}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $2$ by $5$ .

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

Subtract $1$ from $10$ .

$f (\frac{1}{5}) = \frac{9}{5}$

The final answer is $\frac{9}{5}$ .

$\frac{9}{5}$

Step 4

Use the $x$ and $y$ values to find where the minimum occurs.

$(\frac{1}{5}, \frac{9}{5})$

Step 5

Find the Maximum/Minimum Value y=5x^2-2x+2

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