Find the Maximum/Minimum Value h(t)=-4t^2+2t-1

$h (t) = - 4 t^{2} + 2 t - 1$

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 $b$ .

$t = - \frac{2}{2 (- 4)}$

Remove parentheses.

$t = - \frac{2}{2 (- 4)}$

Simplify $- \frac{2}{2 (- 4)}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$t = - \frac{2}{2 \cdot - 4}$

Rewrite the expression.

$t = - \frac{1}{- 4}$

Move the negative in front of the fraction.

$t = - - \frac{1}{4}$

Multiply $- - \frac{1}{4}$ .

Multiply $- 1$ by $- 1$ .

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

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

$t = \frac{1}{4}$

Step 3

Evaluate $f (\frac{1}{4})$ .

Replace the variable $t$ with $\frac{1}{4}$ in the expression.

$h (\frac{1}{4}) = - 4 {(\frac{1}{4})}^{2} + 2 (\frac{1}{4}) - 1$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

$h (\frac{1}{4}) = - 4 \frac{1^{2}}{4^{2}} + 2 (\frac{1}{4}) - 1$

One to any power is one.

$h (\frac{1}{4}) = - 4 \frac{1}{4^{2}} + 2 (\frac{1}{4}) - 1$

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

$h (\frac{1}{4}) = - 4 (\frac{1}{16}) + 2 (\frac{1}{4}) - 1$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$h (\frac{1}{4}) = 4 (- 1) (\frac{1}{16}) + 2 (\frac{1}{4}) - 1$

Factor $4$ out of $16$ .

$h (\frac{1}{4}) = 4 \cdot (- 1 \frac{1}{4 \cdot 4}) + 2 (\frac{1}{4}) - 1$

Cancel the common factor.

$h (\frac{1}{4}) = 4 \cdot (- 1 \frac{1}{4 \cdot 4}) + 2 (\frac{1}{4}) - 1$

Rewrite the expression.

$h (\frac{1}{4}) = - 1 (\frac{1}{4}) + 2 (\frac{1}{4}) - 1$

Rewrite $- 1 (\frac{1}{4})$ as $- (\frac{1}{4})$ .

$h (\frac{1}{4}) = - (\frac{1}{4}) + 2 (\frac{1}{4}) - 1$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$h (\frac{1}{4}) = - \frac{1}{4} + 2 (\frac{1}{2 (2)}) - 1$

Cancel the common factor.

$h (\frac{1}{4}) = - \frac{1}{4} + 2 (\frac{1}{2 \cdot 2}) - 1$

Rewrite the expression.

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{1}{2} - 1$

Find the common denominator.

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{1}{2} \cdot \frac{2}{2} - 1$

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} - 1$

Write $- 1$ as a fraction with denominator $1$ .

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{- 1}{1}$

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

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{- 1}{1} \cdot \frac{4}{4}$

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

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{- 1 \cdot 4}{4}$

Multiply $2$ by $2$ .

$h (\frac{1}{4}) = - \frac{1}{4} + \frac{2}{4} + \frac{- 1 \cdot 4}{4}$

Combine the numerators over the common denominator.

$h (\frac{1}{4}) = \frac{- 1 + 2 - 1 \cdot 4}{4}$

Simplify the expression.

Multiply $- 1$ by $4$ .

$h (\frac{1}{4}) = \frac{- 1 + 2 - 4}{4}$

Add $- 1$ and $2$ .

$h (\frac{1}{4}) = \frac{1 - 4}{4}$

Subtract $4$ from $1$ .

$h (\frac{1}{4}) = \frac{- 3}{4}$

Move the negative in front of the fraction.

$h (\frac{1}{4}) = - \frac{3}{4}$

The final answer is $- \frac{3}{4}$ .

$- \frac{3}{4}$

Step 4

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

$(\frac{1}{4}, - \frac{3}{4})$

Step 5

Find the Maximum/Minimum Value h(t)=-4t^2+2t-1

Related Questions

Support LCMGCF.com by disabling your adblocker.