Find the Maximum/Minimum Value -4.9x^2+24.5x+1

$- 4.9 x^{2} + 24.5 x + 1$

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

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

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

$x = - \frac{24.5}{2 (- 4.9)}$

Remove parentheses.

$x = - \frac{24.5}{2 (- 4.9)}$

Simplify $- \frac{24.5}{2 (- 4.9)}$ .

Multiply $2$ by $- 4.9$ .

$x = - \frac{24.5}{- 9.8}$

Divide $24.5$ by $- 9.8$ .

$x = - - 2.5$

Multiply $- 1$ by $- 2.5$ .

$x = 2.5$

Evaluate $f (2.5)$ .

Replace the variable $x$ with $2.5$ in the expression.

$f (2.5) = - 4.9 {(2.5)}^{2} + 24.5 (2.5) + 1$

Simplify the result.

Simplify each term.

Raise $2.5$ to the power of $2$ .

$f (2.5) = - 4.9 \cdot 6.25 + 24.5 (2.5) + 1$

Multiply $- 4.9$ by $6.25$ .

$f (2.5) = - 30.625 + 24.5 (2.5) + 1$

Multiply $24.5$ by $2.5$ .

$f (2.5) = - 30.625 + 61.25 + 1$

Simplify by adding numbers.

Add $- 30.625$ and $61.25$ .

$f (2.5) = 30.625 + 1$

Add $30.625$ and $1$ .

$f (2.5) = 31.625$

The final answer is $31.625$ .

$31.625$

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

$(2.5, 31.625)$

Find the Maximum/Minimum Value -4.9x^2+24.5x+1

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