Find the Maximum/Minimum Value f(x)=-3x^2+30x-7

$f (x) = - 3 x^{2} + 30 x - 7$

Step 1

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

$f_{max}$ $x = a x^{2} + b x + c$ 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 $15$ .

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

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$x = - (- 1 \cdot 5)$

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

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

$f (5) = - 3 {(5)}^{2} + 30 (5) - 7$

Simplify the result.

Simplify each term.

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

$f (5) = - 3 \cdot 25 + 30 (5) - 7$

Multiply $- 3$ by $25$ .

$f (5) = - 75 + 30 (5) - 7$

Multiply $30$ by $5$ .

$f (5) = - 75 + 150 - 7$

Simplify by adding and subtracting.

Add $- 75$ and $150$ .

$f (5) = 75 - 7$

Subtract $7$ from $75$ .

$f (5) = 68$

The final answer is $68$ .

$68$

Step 4

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

$(5, 68)$

Step 5

Find the Maximum/Minimum Value f(x)=-3x^2+30x-7

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