Find the Maximum/Minimum Value f(x)=-2x^2+4x-3

$f (x) = - 2 x^{2} + 4 x - 3$

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 $x = - \frac{b}{2 a}$

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

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

$x = - \frac{4}{2 (- 2)}$

Remove parentheses.

$x = - \frac{4}{2 (- 2)}$

Simplify $- \frac{4}{2 (- 2)}$ .

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

$x = - \frac{2}{- 2}$

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

Factor $2$ out of $2$ .

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

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

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

Multiply.

Multiply $- 1$ by $1$ .

$x = - - 1$

Multiply $- 1$ by $- 1$ .

$x = 1$

Evaluate $f (1)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 2 {(1)}^{2} + 4 (1) - 3$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = - 2 \cdot 1 + 4 (1) - 3$

Multiply $- 2$ by $1$ .

$f (1) = - 2 + 4 (1) - 3$

Multiply $4$ by $1$ .

$f (1) = - 2 + 4 - 3$

Simplify by adding and subtracting.

Add $- 2$ and $4$ .

$f (1) = 2 - 3$

Subtract $3$ from $2$ .

$f (1) = - 1$

The final answer is $- 1$ .

$- 1$

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

$(1, - 1)$

Find the Maximum/Minimum Value f(x)=-2x^2+4x-3

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