Find the Maximum/Minimum Value y=5x^2-30x+47

$y = 5 x^{2} - 30 x + 47$

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 $b$ .

$x = - \frac{- 30}{2 (5)}$

Remove parentheses.

$x = - \frac{- 30}{2 (5)}$

Simplify $- \frac{- 30}{2 (5)}$ .

Cancel the common factor of $- 30$ and $2$ .

Factor $2$ out of $- 30$ .

$x = - \frac{2 \cdot - 15}{2 \cdot 5}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 15}{2 (5)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 15}{2 \cdot 5}$

Rewrite the expression.

$x = - \frac{- 15}{5}$

Cancel the common factor of $- 15$ and $5$ .

Factor $5$ out of $- 15$ .

$x = - \frac{5 \cdot - 3}{5}$

Cancel the common factors.

Factor $5$ out of $5$ .

$x = - \frac{5 \cdot - 3}{5 (1)}$

Cancel the common factor.

$x = - \frac{5 \cdot - 3}{5 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 3}{1}$

Divide $- 3$ by $1$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = 5 {(3)}^{2} - 30 \cdot 3 + 47$

Simplify the result.

Simplify each term.

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

$f (3) = 5 \cdot 9 - 30 \cdot 3 + 47$

Multiply $5$ by $9$ .

$f (3) = 45 - 30 \cdot 3 + 47$

Multiply $- 30$ by $3$ .

$f (3) = 45 - 90 + 47$

Simplify by adding and subtracting.

Subtract $90$ from $45$ .

$f (3) = - 45 + 47$

Add $- 45$ and $47$ .

$f (3) = 2$

The final answer is $2$ .

$2$

Step 4

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

$(3, 2)$

Step 5

Find the Maximum/Minimum Value y=5x^2-30x+47

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