Find the Maximum/Minimum Value y=-5x^2+10x+12

$y = - 5 x^{2} + 10 x + 12$

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 $b$ .

$x = - \frac{10}{2 (- 5)}$

Remove parentheses.

$x = - \frac{10}{2 (- 5)}$

Simplify $- \frac{10}{2 (- 5)}$ .

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$x = - \frac{2 \cdot 5}{2 \cdot - 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 5$ .

$x = - \frac{2 \cdot 5}{2 (- 5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 5}{2 \cdot - 5}$

Rewrite the expression.

$x = - \frac{5}{- 5}$

Cancel the common factor of $5$ and $- 5$ .

Factor $5$ out of $5$ .

$x = - \frac{5 (1)}{- 5}$

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

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

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

Multiply $- 1$ by $1$ .

$x = - - 1$

Multiply $- 1$ by $- 1$ .

$x = 1$

Step 3

Evaluate $f (1)$ .

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

$f (1) = - 5 {(1)}^{2} + 10 (1) + 12$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = - 5 \cdot 1 + 10 (1) + 12$

Multiply $- 5$ by $1$ .

$f (1) = - 5 + 10 (1) + 12$

Multiply $10$ by $1$ .

$f (1) = - 5 + 10 + 12$

Simplify by adding numbers.

Add $- 5$ and $10$ .

$f (1) = 5 + 12$

Add $5$ and $12$ .

$f (1) = 17$

The final answer is $17$ .

$17$

Step 4

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

$(1, 17)$

Step 5

Find the Maximum/Minimum Value y=-5x^2+10x+12

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