Find the Maximum/Minimum Value 4x^2-9x-7

$4 x^{2} - 9 x - 7$

Step 1

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

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

Step 2

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

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

$x = - \frac{- 9}{2 (4)}$

Remove parentheses.

$x = - \frac{- 9}{2 (4)}$

Simplify $- \frac{- 9}{2 (4)}$ .

Multiply $2$ by $4$ .

$x = - \frac{- 9}{8}$

Move the negative in front of the fraction.

$x = - - \frac{9}{8}$

Multiply $- - \frac{9}{8}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{9}{8})$

Multiply $\frac{9}{8}$ by $1$ .

$x = \frac{9}{8}$

Step 3

Evaluate $f (\frac{9}{8})$ .

Replace the variable $x$ with $\frac{9}{8}$ in the expression.

$f (\frac{9}{8}) = 4 {(\frac{9}{8})}^{2} - 9 (\frac{9}{8}) - 7$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{9}{8}$ .

$f (\frac{9}{8}) = 4 (\frac{9^{2}}{8^{2}}) - 9 (\frac{9}{8}) - 7$

Raise $9$ to the power of $2$ .

$f (\frac{9}{8}) = 4 (\frac{81}{8^{2}}) - 9 (\frac{9}{8}) - 7$

Raise $8$ to the power of $2$ .

$f (\frac{9}{8}) = 4 (\frac{81}{64}) - 9 (\frac{9}{8}) - 7$

Cancel the common factor of $4$ .

Factor $4$ out of $64$ .

$f (\frac{9}{8}) = 4 (\frac{81}{4 (16)}) - 9 (\frac{9}{8}) - 7$

Cancel the common factor.

$f (\frac{9}{8}) = 4 (\frac{81}{4 \cdot 16}) - 9 (\frac{9}{8}) - 7$

Rewrite the expression.

$f (\frac{9}{8}) = \frac{81}{16} - 9 (\frac{9}{8}) - 7$

Multiply $- 9 (\frac{9}{8})$ .

Combine $- 9$ and $\frac{9}{8}$ .

$f (\frac{9}{8}) = \frac{81}{16} + \frac{- 9 \cdot 9}{8} - 7$

Multiply $- 9$ by $9$ .

$f (\frac{9}{8}) = \frac{81}{16} + \frac{- 81}{8} - 7$

Move the negative in front of the fraction.

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81}{8} - 7$

Find the common denominator.

Multiply $\frac{81}{8}$ by $\frac{2}{2}$ .

$f (\frac{9}{8}) = \frac{81}{16} - (\frac{81}{8} \cdot \frac{2}{2}) - 7$

Multiply $\frac{81}{8}$ by $\frac{2}{2}$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{8 \cdot 2} - 7$

Write $- 7$ as a fraction with denominator $1$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{8 \cdot 2} + \frac{- 7}{1}$

Multiply $\frac{- 7}{1}$ by $\frac{16}{16}$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{8 \cdot 2} + \frac{- 7}{1} \cdot \frac{16}{16}$

Multiply $\frac{- 7}{1}$ by $\frac{16}{16}$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{8 \cdot 2} + \frac{- 7 \cdot 16}{16}$

Reorder the factors of $8 \cdot 2$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{2 \cdot 8} + \frac{- 7 \cdot 16}{16}$

Multiply $2$ by $8$ .

$f (\frac{9}{8}) = \frac{81}{16} - \frac{81 \cdot 2}{16} + \frac{- 7 \cdot 16}{16}$

Combine the numerators over the common denominator.

$f (\frac{9}{8}) = \frac{81 - 81 \cdot 2 - 7 \cdot 16}{16}$

Simplify each term.

Multiply $- 81$ by $2$ .

$f (\frac{9}{8}) = \frac{81 - 162 - 7 \cdot 16}{16}$

Multiply $- 7$ by $16$ .

$f (\frac{9}{8}) = \frac{81 - 162 - 112}{16}$

Simplify the expression.

Subtract $162$ from $81$ .

$f (\frac{9}{8}) = \frac{- 81 - 112}{16}$

Subtract $112$ from $- 81$ .

$f (\frac{9}{8}) = \frac{- 193}{16}$

Move the negative in front of the fraction.

$f (\frac{9}{8}) = - \frac{193}{16}$

The final answer is $- \frac{193}{16}$ .

$- \frac{193}{16}$

Step 4

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

$(\frac{9}{8}, - \frac{193}{16})$

Step 5

Find the Maximum/Minimum Value 4x^2-9x-7

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