Find the Maximum/Minimum Value p(x)=-3(x-5)^2+12

$p (x) = - 3 {(x - 5)}^{2} + 12$

Step 1

Simplify each term.

Rewrite ${(x - 5)}^{2}$ as $(x - 5) (x - 5)$ .

$p (x) = - 3 ((x - 5) (x - 5)) + 12$

Expand $(x - 5) (x - 5)$ using the FOIL Method.

Apply the distributive property.

$p (x) = - 3 (x (x - 5) - 5 (x - 5)) + 12$

Apply the distributive property.

$p (x) = - 3 (x \cdot x + x \cdot - 5 - 5 (x - 5)) + 12$

Apply the distributive property.

$p (x) = - 3 (x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5) + 12$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$p (x) = - 3 (x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5) + 12$

Move $- 5$ to the left of $x$ .

$p (x) = - 3 (x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5) + 12$

Multiply $- 5$ by $- 5$ .

$p (x) = - 3 (x^{2} - 5 x - 5 x + 25) + 12$

Subtract $5 x$ from $- 5 x$ .

$p (x) = - 3 (x^{2} - 10 x + 25) + 12$

Apply the distributive property.

$p (x) = - 3 x^{2} - 3 (- 10 x) - 3 \cdot 25 + 12$

Simplify.

Multiply $- 10$ by $- 3$ .

$p (x) = - 3 x^{2} + 30 x - 3 \cdot 25 + 12$

Multiply $- 3$ by $25$ .

$p (x) = - 3 x^{2} + 30 x - 75 + 12$

Step 2

Add $- 75$ and $12$ .

$p (x) = - 3 x^{2} + 30 x - 63$

Step 3

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 4

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

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

$x = - \frac{30}{2 (- 3)}$

Remove parentheses.

$x = - \frac{30}{2 (- 3)}$

Simplify $- \frac{30}{2 (- 3)}$ .

Cancel the common factor of $30$ and $2$ .

Factor $2$ out of $30$ .

$x = - \frac{2 \cdot 15}{2 \cdot - 3}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 15}{2 (- 3)}$

Cancel the common factor.

$x = - \frac{2 \cdot 15}{2 \cdot - 3}$

Rewrite the expression.

$x = - \frac{15}{- 3}$

Cancel the common factor of $15$ and $- 3$ .

Factor $3$ out of $15$ .

$x = - \frac{3 (5)}{- 3}$

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$x = - (- 1 \cdot 5)$

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 5

Evaluate $f (5)$ .

Replace the variable $x$ with $5$ in the expression.

$p (5) = - 3 {(5)}^{2} + 30 (5) - 63$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$p (5) = - 3 \cdot 25 + 30 (5) - 63$

Multiply $- 3$ by $25$ .

$p (5) = - 75 + 30 (5) - 63$

Multiply $30$ by $5$ .

$p (5) = - 75 + 150 - 63$

Simplify by adding and subtracting.

Add $- 75$ and $150$ .

$p (5) = 75 - 63$

Subtract $63$ from $75$ .

$p (5) = 12$

The final answer is $12$ .

$12$

Step 6

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

$(5, 12)$

Step 7

Find the Maximum/Minimum Value p(x)=-3(x-5)^2+12

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