Algebra Examples

$h (t) = - 16 t^{2} + 24 t + 6$

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 $t = - \frac{b}{2 a}$ .

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Step 2.1

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

$t = - \frac{24}{2 (- 16)}$

Step 2.2

Remove parentheses.

$t = - \frac{24}{2 (- 16)}$

Step 2.3

Simplify $- \frac{24}{2 (- 16)}$ .

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Step 2.3.1

Cancel the common factor of $24$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $24$ .

$t = - \frac{2 \cdot 12}{2 \cdot - 16}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $2 \cdot - 16$ .

$t = - \frac{2 \cdot 12}{2 (- 16)}$

Step 2.3.1.2.2

Cancel the common factor.

$t = - \frac{2 \cdot 12}{2 \cdot - 16}$

Step 2.3.1.2.3

Rewrite the expression.

$t = - \frac{12}{- 16}$

Step 2.3.2

Cancel the common factor of $12$ and $- 16$ .

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Step 2.3.2.1

Factor $4$ out of $12$ .

$t = - \frac{4 (3)}{- 16}$

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 3}{4 \cdot - 4}$

Step 2.3.2.2.2

Cancel the common factor.

$t = - \frac{4 \cdot 3}{4 \cdot - 4}$

Step 2.3.2.2.3

Rewrite the expression.

$t = - \frac{3}{- 4}$

Step 2.3.3

Move the negative in front of the fraction.

$t = - - \frac{3}{4}$

Step 2.3.4

Multiply $- - \frac{3}{4}$ .

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Step 2.3.4.1

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{3}{4})$

Step 2.3.4.2

Multiply $\frac{3}{4}$ by $1$ .

$t = \frac{3}{4}$

Step 3

Evaluate $f (\frac{3}{4})$ .

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Step 3.1

Replace the variable $t$ with $\frac{3}{4}$ in the expression.

$h (\frac{3}{4}) = - 16 {(\frac{3}{4})}^{2} + 24 (\frac{3}{4}) + 6$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to $\frac{3}{4}$ .

$h (\frac{3}{4}) = - 16 \frac{3^{2}}{4^{2}} + 24 (\frac{3}{4}) + 6$

Step 3.2.1.2

Raise $3$ to the power of $2$ .

$h (\frac{3}{4}) = - 16 \frac{9}{4^{2}} + 24 (\frac{3}{4}) + 6$

Step 3.2.1.3

Raise $4$ to the power of $2$ .

$h (\frac{3}{4}) = - 16 (\frac{9}{16}) + 24 (\frac{3}{4}) + 6$

Step 3.2.1.4

Cancel the common factor of $16$ .

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Step 3.2.1.4.1

Factor $16$ out of $- 16$ .

$h (\frac{3}{4}) = 16 (- 1) (\frac{9}{16}) + 24 (\frac{3}{4}) + 6$

Step 3.2.1.4.2

Cancel the common factor.

$h (\frac{3}{4}) = 16 \cdot (- 1 (\frac{9}{16})) + 24 (\frac{3}{4}) + 6$

Step 3.2.1.4.3

Rewrite the expression.

$h (\frac{3}{4}) = - 1 \cdot 9 + 24 (\frac{3}{4}) + 6$

Step 3.2.1.5

Multiply $- 1$ by $9$ .

$h (\frac{3}{4}) = - 9 + 24 (\frac{3}{4}) + 6$

Step 3.2.1.6

Cancel the common factor of $4$ .

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Step 3.2.1.6.1

Factor $4$ out of $24$ .

$h (\frac{3}{4}) = - 9 + 4 (6) (\frac{3}{4}) + 6$

Step 3.2.1.6.2

Cancel the common factor.

$h (\frac{3}{4}) = - 9 + 4 \cdot (6 (\frac{3}{4})) + 6$

Step 3.2.1.6.3

Rewrite the expression.

$h (\frac{3}{4}) = - 9 + 6 \cdot 3 + 6$

Step 3.2.1.7

Multiply $6$ by $3$ .

$h (\frac{3}{4}) = - 9 + 18 + 6$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $- 9$ and $18$ .

$h (\frac{3}{4}) = 9 + 6$

Step 3.2.2.2

Add $9$ and $6$ .

$h (\frac{3}{4}) = 15$

Step 3.2.3

The final answer is $15$ .

$15$

Step 4

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

$(\frac{3}{4}, 15)$

Step 5