Find the Maximum/Minimum Value y=3x^2+8x+7

$y = 3 x^{2} + 8 x + 7$

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 $x = - \frac{b}{2 a}$

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

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

$x = - \frac{8}{2 (3)}$

Remove parentheses.

$x = - \frac{8}{2 (3)}$

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$f (- \frac{4}{3}) = 3 {(- \frac{4}{3})}^{2} + 8 (- \frac{4}{3}) + 7$

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$f (- \frac{4}{3}) = 3 ({(- 1)}^{2} {(\frac{4}{3})}^{2}) + 8 (- \frac{4}{3}) + 7$

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

$f (- \frac{4}{3}) = 3 ({(- 1)}^{2} (\frac{4^{2}}{3^{2}})) + 8 (- \frac{4}{3}) + 7$

Raise $- 1$ to the power of $2$ .

$f (- \frac{4}{3}) = 3 (1 (\frac{4^{2}}{3^{2}})) + 8 (- \frac{4}{3}) + 7$

Multiply $\frac{4^{2}}{3^{2}}$ by $1$ .

$f (- \frac{4}{3}) = 3 (\frac{4^{2}}{3^{2}}) + 8 (- \frac{4}{3}) + 7$

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

$f (- \frac{4}{3}) = 3 (\frac{16}{3^{2}}) + 8 (- \frac{4}{3}) + 7$

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

$f (- \frac{4}{3}) = 3 (\frac{16}{9}) + 8 (- \frac{4}{3}) + 7$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$f (- \frac{4}{3}) = 3 (\frac{16}{3 (3)}) + 8 (- \frac{4}{3}) + 7$

Cancel the common factor.

$f (- \frac{4}{3}) = 3 (\frac{16}{3 \cdot 3}) + 8 (- \frac{4}{3}) + 7$

Rewrite the expression.

$f (- \frac{4}{3}) = \frac{16}{3} + 8 (- \frac{4}{3}) + 7$

Multiply $8 (- \frac{4}{3})$ .

Multiply $- 1$ by $8$ .

$f (- \frac{4}{3}) = \frac{16}{3} - 8 (\frac{4}{3}) + 7$

Combine $- 8$ and $\frac{4}{3}$ .

$f (- \frac{4}{3}) = \frac{16}{3} + \frac{- 8 \cdot 4}{3} + 7$

Multiply $- 8$ by $4$ .

$f (- \frac{4}{3}) = \frac{16}{3} + \frac{- 32}{3} + 7$

Move the negative in front of the fraction.

$f (- \frac{4}{3}) = \frac{16}{3} - \frac{32}{3} + 7$

Combine fractions.

Combine the numerators over the common denominator.

$f (- \frac{4}{3}) = 7 + \frac{16 - 32}{3}$

Simplify the expression.

Subtract $32$ from $16$ .

$f (- \frac{4}{3}) = 7 + \frac{- 16}{3}$

Move the negative in front of the fraction.

$f (- \frac{4}{3}) = 7 - \frac{16}{3}$

To write $7$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (- \frac{4}{3}) = 7 \cdot \frac{3}{3} - \frac{16}{3}$

Combine $7$ and $\frac{3}{3}$ .

$f (- \frac{4}{3}) = \frac{7 \cdot 3}{3} - \frac{16}{3}$

Combine the numerators over the common denominator.

$f (- \frac{4}{3}) = \frac{7 \cdot 3 - 16}{3}$

Simplify the numerator.

Multiply $7$ by $3$ .

$f (- \frac{4}{3}) = \frac{21 - 16}{3}$

Subtract $16$ from $21$ .

$f (- \frac{4}{3}) = \frac{5}{3}$

The final answer is $\frac{5}{3}$ .

$\frac{5}{3}$

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

$(- \frac{4}{3}, \frac{5}{3})$

Find the Maximum/Minimum Value y=3x^2+8x+7

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