Find the Maximum/Minimum Value y=-x^2+2x+3

$y = - x^{2} + 2 x + 3$

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 $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \frac{2}{2 \cdot - 1}$

Rewrite the expression.

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

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.

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

Simplify the result.

Simplify each term.

One to any power is one.

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

Multiply $- 1$ by $1$ .

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

Multiply $2$ by $1$ .

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

Simplify by adding numbers.

Add $- 1$ and $2$ .

$f (1) = 1 + 3$

Add $1$ and $3$ .

$f (1) = 4$

The final answer is $4$ .

$4$

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

$(1, 4)$

Find the Maximum/Minimum Value y=-x^2+2x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.