Find the Maximum/Minimum Value (x+1)^2-4

${(x + 1)}^{2} - 4$

Step 1

Simplify each term.

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$f (x) = (x + 1) (x + 1) - 4$

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$f (x) = x (x + 1) + 1 (x + 1) - 4$

Apply the distributive property.

$f (x) = x \cdot x + x \cdot 1 + 1 (x + 1) - 4$

Apply the distributive property.

$f (x) = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1 - 4$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$f (x) = x^{2} + x \cdot 1 + 1 x + 1 \cdot 1 - 4$

Multiply $x$ by $1$ .

$f (x) = x^{2} + x + 1 x + 1 \cdot 1 - 4$

Multiply $x$ by $1$ .

$f (x) = x^{2} + x + x + 1 \cdot 1 - 4$

Multiply $1$ by $1$ .

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

Add $x$ and $x$ .

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

Step 2

Subtract $4$ from $1$ .

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

Step 3

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 4

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.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 5

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.

Raise $- 1$ to the power of $2$ .

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

Multiply $2$ by $- 1$ .

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

Simplify by subtracting numbers.

Subtract $2$ from $1$ .

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

Subtract $3$ from $- 1$ .

$f (- 1) = - 4$

The final answer is $- 4$ .

$- 4$

Step 6

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

$(- 1, - 4)$

Step 7

Find the Maximum/Minimum Value (x+1)^2-4

Related Questions

Support LCMGCF.com by disabling your adblocker.