Find the Maximum/Minimum Value h(t)=-4.9t^2+14.7t+1

$h (t) = - 4.9 t^{2} + 14.7 t + 1$

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 $b$ .

$t = - \frac{14.7}{2 (- 4.9)}$

Remove parentheses.

$t = - \frac{14.7}{2 (- 4.9)}$

Simplify $- \frac{14.7}{2 (- 4.9)}$ .

Multiply $2$ by $- 4.9$ .

$t = - \frac{14.7}{- 9.8}$

Divide $14.7$ by $- 9.8$ .

$t = - - 1.5$

Multiply $- 1$ by $- 1.5$ .

$t = 1.5$

Step 3

Evaluate $f (1.5)$ .

Replace the variable $t$ with $1.5$ in the expression.

$h (1.5) = - 4.9 {(1.5)}^{2} + 14.7 (1.5) + 1$

Simplify the result.

Simplify each term.

Raise $1.5$ to the power of $2$ .

$h (1.5) = - 4.9 \cdot 2.25 + 14.7 (1.5) + 1$

Multiply $- 4.9$ by $2.25$ .

$h (1.5) = - 11.025 + 14.7 (1.5) + 1$

Multiply $14.7$ by $1.5$ .

$h (1.5) = - 11.025 + 22.05 + 1$

Simplify by adding numbers.

Add $- 11.025$ and $22.05$ .

$h (1.5) = 11.025 + 1$

Add $11.025$ and $1$ .

$h (1.5) = 12.025$

The final answer is $12.025$ .

$12.025$

Step 4

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

$(1.5, 12.025)$

Step 5

Find the Maximum/Minimum Value h(t)=-4.9t^2+14.7t+1

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