Find the Maximum/Minimum Value f(x)=3x^2-30x+79

$f (x) = 3 x^{2} - 30 x + 79$

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 $b$ .

$x = - \frac{- 30}{2 (3)}$

Remove parentheses.

$x = - \frac{- 30}{2 (3)}$

Simplify $- \frac{- 30}{2 (3)}$ .

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

Factor $2$ out of $- 30$ .

$x = - \frac{2 \cdot - 15}{2 \cdot 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 3$ .

$x = - \frac{2 \cdot - 15}{2 (3)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 15}{2 \cdot 3}$

Rewrite the expression.

$x = - \frac{- 15}{3}$

Cancel the common factor of $- 15$ and $3$ .

Factor $3$ out of $- 15$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 5$ by $1$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

Replace the variable $x$ with $5$ in the expression.

$f (5) = 3 {(5)}^{2} - 30 \cdot 5 + 79$

Simplify the result.

Simplify each term.

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

$f (5) = 3 \cdot 25 - 30 \cdot 5 + 79$

Multiply $3$ by $25$ .

$f (5) = 75 - 30 \cdot 5 + 79$

Multiply $- 30$ by $5$ .

$f (5) = 75 - 150 + 79$

Simplify by adding and subtracting.

Subtract $150$ from $75$ .

$f (5) = - 75 + 79$

Add $- 75$ and $79$ .

$f (5) = 4$

The final answer is $4$ .

$4$

Step 4

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

$(5, 4)$

Step 5

Find the Maximum/Minimum Value f(x)=3x^2-30x+79

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