Find the Maximum/Minimum Value h(t)=-4.9t^2+9.8t+1

$h (t) = - 4.9 t^{2} + 9.8 t + 1$

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 $b$ .

$t = - \frac{9.8}{2 (- 4.9)}$

Remove parentheses.

$t = - \frac{9.8}{2 (- 4.9)}$

Simplify $- \frac{9.8}{2 (- 4.9)}$ .

Multiply $2$ by $- 4.9$ .

$t = - \frac{9.8}{- 9.8}$

Divide $9.8$ by $- 9.8$ .

$t = - - 1$

Multiply $- 1$ by $- 1$ .

$t = 1$

Step 3

Evaluate $f (1)$ .

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

$h (1) = - 4.9 {(1)}^{2} + 9.8 (1) + 1$

Simplify the result.

Simplify each term.

One to any power is one.

$h (1) = - 4.9 \cdot 1 + 9.8 (1) + 1$

Multiply $- 4.9$ by $1$ .

$h (1) = - 4.9 + 9.8 (1) + 1$

Multiply $9.8$ by $1$ .

$h (1) = - 4.9 + 9.8 + 1$

Simplify by adding numbers.

Add $- 4.9$ and $9.8$ .

$h (1) = 4.9 + 1$

Add $4.9$ and $1$ .

$h (1) = 5.9$

The final answer is $5.9$ .

$5.9$

Step 4

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

$(1, 5.9)$

Step 5

Find the Maximum/Minimum Value h(t)=-4.9t^2+9.8t+1

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