Find the Vertex f(x)=(x+3)(x-5)

$f (x) = (x + 3) (x - 5)$

Step 1

Rewrite the equation in vertex form.

Complete the square for $(x + 3) (x - 5)$ .

Simplify the expression.

Expand $(x + 3) (x - 5)$ using the FOIL Method.

Apply the distributive property.

$x (x - 5) + 3 (x - 5)$

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 3 (x - 5)$

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 3 x + 3 \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 + 3 x + 3 \cdot - 5$

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

$x^{2} - 5 \cdot x + 3 x + 3 \cdot - 5$

Multiply $3$ by $- 5$ .

$x^{2} - 5 x + 3 x - 15$

Add $- 5 x$ and $3 x$ .

$x^{2} - 2 x - 15$

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 2$

$c = - 15$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 2}{2 \cdot 1}$

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

Factor $2$ out of $- 2$ .

$d = \frac{2 \cdot - 1}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 1}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 1}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 1}{1}$

Divide $- 1$ by $1$ .

$d = - 1$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 15 - \frac{{(- 2)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = - 15 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 15 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = - 15 - \frac{4}{4}$

Rewrite the expression.

$e = - 15 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = - 15 - 1$

Subtract $1$ from $- 15$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 1)}^{2} - 16$ .

${(x - 1)}^{2} - 16$

Set $y$ equal to the new right side.

$y = {(x - 1)}^{2} - 16$

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 1$

$h = 1$

$k = - 16$

Step 3

Find the vertex $(h, k)$ .

$(1, - 16)$

Step 4

Find the Vertex f(x)=(x+3)(x-5)

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