Find the Maximum/Minimum Value -4.9t^2+50t+2

$- 4.9 t^{2} + 50 t + 2$

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 $f (- \frac{b}{2 a})$ .

$f_{max}$ $t = a t^{2} + b t + c$ occurs at $t = - \frac{b}{2 a}$

Step 2

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

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

$t = - \frac{50}{2 (- 4.9)}$

Remove parentheses.

$t = - \frac{50}{2 (- 4.9)}$

Simplify $- \frac{50}{2 (- 4.9)}$ .

Cancel the common factor of $50$ and $2$ .

Factor $2$ out of $50$ .

$t = - \frac{2 \cdot 25}{2 \cdot - 4.9}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4.9$ .

$t = - \frac{2 \cdot 25}{2 (- 4.9)}$

Cancel the common factor.

$t = - \frac{2 \cdot 25}{2 \cdot - 4.9}$

Rewrite the expression.

$t = - \frac{25}{- 4.9}$

Simplify the expression.

Divide $25$ by $- 4.9$ .

$t = - - 5.10204081$

Multiply $- 1$ by $- 5.10204081$ .

$t = 5.10204081$

Step 3

Evaluate $f (5.10204081)$ .

Replace the variable $t$ with $5.10204081$ in the expression.

$f (5.10204081) = - 4.9 {(5.10204081)}^{2} + 50 (5.10204081) + 2$

Simplify the result.

Simplify each term.

Raise $5.10204081$ to the power of $2$ .

$f (5.10204081) = - 4.9 \cdot 26.03082049 + 50 (5.10204081) + 2$

Multiply $- 4.9$ by $26.03082049$ .

$f (5.10204081) = - 127.5510204 + 50 (5.10204081) + 2$

Multiply $50$ by $5.10204081$ .

$f (5.10204081) = - 127.5510204 + 255.10204081 + 2$

Simplify by adding numbers.

Add $- 127.5510204$ and $255.10204081$ .

$f (5.10204081) = 127.5510204 + 2$

Add $127.5510204$ and $2$ .

$f (5.10204081) = 129.5510204$

The final answer is $129.5510204$ .

$129.5510204$

Step 4

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

$(5.10204081, 129.5510204)$

Step 5

Find the Maximum/Minimum Value -4.9t^2+50t+2

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