Find the Maximum/Minimum Value y=1/5x^2-5x+27

$y = \frac{1}{5} x^{2} - 5 x + 27$

Step 1

Combine $\frac{1}{5}$ and $x^{2}$ .

$f (x) = \frac{x^{2}}{5} - 5 x + 27$

Step 2

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 3

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

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

$x = - \frac{- 5}{2 (0.2)}$

Remove parentheses.

$x = - \frac{- 5}{2 (0.2)}$

Simplify $- \frac{- 5}{2 (0.2)}$ .

Multiply $2$ by $0.2$ .

$x = - \frac{- 5}{0.4}$

Divide $- 5$ by $0.4$ .

$x = - - 12.5$

Multiply $- 1$ by $- 12.5$ .

$x = 12.5$

Step 4

Evaluate $f (12.5)$ .

Replace the variable $x$ with $12.5$ in the expression.

$f (12.5) = \frac{{(12.5)}^{2}}{5} - 5 \cdot 12.5 + 27$

Simplify the result.

Simplify each term.

Raise $12.5$ to the power of $2$ .

$f (12.5) = \frac{156.25}{5} - 5 \cdot 12.5 + 27$

Divide $156.25$ by $5$ .

$f (12.5) = 31.25 - 5 \cdot 12.5 + 27$

Multiply $- 5$ by $12.5$ .

$f (12.5) = 31.25 - 62.5 + 27$

Simplify by adding and subtracting.

Subtract $62.5$ from $31.25$ .

$f (12.5) = - 31.25 + 27$

Add $- 31.25$ and $27$ .

$f (12.5) = - 4.25$

The final answer is $- 4.25$ .

$- 4.25$

Step 5

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

$(12.5, - 4.25)$

Step 6

Find the Maximum/Minimum Value y=1/5x^2-5x+27

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