Find the Maximum/Minimum Value f(x)=-4x^2+6x+3

$f (x) = - 4 x^{2} + 6 x + 3$

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 $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ 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 $b$ .

$x = - \frac{6}{2 (- 4)}$

Remove parentheses.

$x = - \frac{6}{2 (- 4)}$

Simplify $- \frac{6}{2 (- 4)}$ .

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot - 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4$ .

$x = - \frac{2 \cdot 3}{2 (- 4)}$

Cancel the common factor.

$x = - \frac{2 \cdot 3}{2 \cdot - 4}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

Step 3

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

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

$f (\frac{3}{4}) = - 4 {(\frac{3}{4})}^{2} + 6 (\frac{3}{4}) + 3$

Simplify the result.

Simplify each term.

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

$f (\frac{3}{4}) = - 4 \frac{3^{2}}{4^{2}} + 6 (\frac{3}{4}) + 3$

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

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

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

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

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

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

Factor $4$ out of $16$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f (\frac{3}{4}) = - 1 (\frac{9}{4}) + 6 (\frac{3}{4}) + 3$

Rewrite $- 1 (\frac{9}{4})$ as $- (\frac{9}{4})$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$f (\frac{3}{4}) = - \frac{9}{4} + 2 (3) (\frac{3}{4}) + 3$

Factor $2$ out of $4$ .

$f (\frac{3}{4}) = - \frac{9}{4} + 2 \cdot (3 (\frac{3}{2 \cdot 2})) + 3$

Cancel the common factor.

$f (\frac{3}{4}) = - \frac{9}{4} + 2 \cdot (3 (\frac{3}{2 \cdot 2})) + 3$

Rewrite the expression.

$f (\frac{3}{4}) = - \frac{9}{4} + 3 (\frac{3}{2}) + 3$

Combine $3$ and $\frac{3}{2}$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{3 \cdot 3}{2} + 3$

Multiply $3$ by $3$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9}{2} + 3$

Find the common denominator.

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9}{2} \cdot \frac{2}{2} + 3$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + 3$

Write $3$ as a fraction with denominator $1$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{3}{1}$

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

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{3}{1} \cdot \frac{4}{4}$

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

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{3 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (\frac{3}{4}) = - \frac{9}{4} + \frac{9 \cdot 2}{4} + \frac{3 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (\frac{3}{4}) = \frac{- 9 + 9 \cdot 2 + 3 \cdot 4}{4}$

Simplify each term.

Multiply $9$ by $2$ .

$f (\frac{3}{4}) = \frac{- 9 + 18 + 3 \cdot 4}{4}$

Multiply $3$ by $4$ .

$f (\frac{3}{4}) = \frac{- 9 + 18 + 12}{4}$

Simplify by adding numbers.

Add $- 9$ and $18$ .

$f (\frac{3}{4}) = \frac{9 + 12}{4}$

Add $9$ and $12$ .

$f (\frac{3}{4}) = \frac{21}{4}$

The final answer is $\frac{21}{4}$ .

$\frac{21}{4}$

Step 4

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

$(\frac{3}{4}, \frac{21}{4})$

Step 5

Find the Maximum/Minimum Value f(x)=-4x^2+6x+3

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