Find the Maximum/Minimum Value y=-10x^2+x-10

$y = - 10 x^{2} + x - 10$

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 $b$ .

$x = - \frac{1}{2 (- 10)}$

Remove parentheses.

$x = - \frac{1}{2 (- 10)}$

Simplify $- \frac{1}{2 (- 10)}$ .

Multiply $2$ by $- 10$ .

$x = - \frac{1}{- 20}$

Move the negative in front of the fraction.

$x = - - \frac{1}{20}$

Multiply $- - \frac{1}{20}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{1}{20})$

Multiply $\frac{1}{20}$ by $1$ .

$x = \frac{1}{20}$

Step 3

Evaluate $f (\frac{1}{20})$ .

Replace the variable $x$ with $\frac{1}{20}$ in the expression.

$f (\frac{1}{20}) = - 10 {(\frac{1}{20})}^{2} + \frac{1}{20} - 10$

Simplify the result.

Remove parentheses.

$f (\frac{1}{20}) = - 10 {(\frac{1}{20})}^{2} + \frac{1}{20} - 10$

Simplify each term.

Apply the product rule to $\frac{1}{20}$ .

$f (\frac{1}{20}) = - 10 \frac{1^{2}}{20^{2}} + \frac{1}{20} - 10$

One to any power is one.

$f (\frac{1}{20}) = - 10 \frac{1}{20^{2}} + \frac{1}{20} - 10$

Raise $20$ to the power of $2$ .

$f (\frac{1}{20}) = - 10 (\frac{1}{400}) + \frac{1}{20} - 10$

Cancel the common factor of $10$ .

Factor $10$ out of $- 10$ .

$f (\frac{1}{20}) = 10 (- 1) (\frac{1}{400}) + \frac{1}{20} - 10$

Factor $10$ out of $400$ .

$f (\frac{1}{20}) = 10 \cdot (- 1 \frac{1}{10 \cdot 40}) + \frac{1}{20} - 10$

Cancel the common factor.

$f (\frac{1}{20}) = 10 \cdot (- 1 \frac{1}{10 \cdot 40}) + \frac{1}{20} - 10$

Rewrite the expression.

$f (\frac{1}{20}) = - 1 (\frac{1}{40}) + \frac{1}{20} - 10$

Rewrite $- 1 (\frac{1}{40})$ as $- (\frac{1}{40})$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{1}{20} - 10$

Find the common denominator.

Multiply $\frac{1}{20}$ by $\frac{2}{2}$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{1}{20} \cdot \frac{2}{2} - 10$

Multiply $\frac{1}{20}$ by $\frac{2}{2}$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{20 \cdot 2} - 10$

Write $- 10$ as a fraction with denominator $1$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{20 \cdot 2} + \frac{- 10}{1}$

Multiply $\frac{- 10}{1}$ by $\frac{40}{40}$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{20 \cdot 2} + \frac{- 10}{1} \cdot \frac{40}{40}$

Multiply $\frac{- 10}{1}$ by $\frac{40}{40}$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{20 \cdot 2} + \frac{- 10 \cdot 40}{40}$

Reorder the factors of $20 \cdot 2$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{2 \cdot 20} + \frac{- 10 \cdot 40}{40}$

Multiply $2$ by $20$ .

$f (\frac{1}{20}) = - \frac{1}{40} + \frac{2}{40} + \frac{- 10 \cdot 40}{40}$

Combine the numerators over the common denominator.

$f (\frac{1}{20}) = \frac{- 1 + 2 - 10 \cdot 40}{40}$

Simplify the expression.

Multiply $- 10$ by $40$ .

$f (\frac{1}{20}) = \frac{- 1 + 2 - 400}{40}$

Add $- 1$ and $2$ .

$f (\frac{1}{20}) = \frac{1 - 400}{40}$

Subtract $400$ from $1$ .

$f (\frac{1}{20}) = \frac{- 399}{40}$

Move the negative in front of the fraction.

$f (\frac{1}{20}) = - \frac{399}{40}$

The final answer is $- \frac{399}{40}$ .

$- \frac{399}{40}$

Step 4

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

$(\frac{1}{20}, - \frac{399}{40})$

Step 5

Find the Maximum/Minimum Value y=-10x^2+x-10

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