Find Three Ordered Pair Solutions h(x)=(x-6)^2-1

$h (x) = {(x - 6)}^{2} - 1$

Step 1

Write $h (x) = {(x - 6)}^{2} - 1$ as an equation.

$y = {(x - 6)}^{2} - 1$

Step 2

Simplify ${(x - 6)}^{2} - 1$ .

Simplify each term.

Rewrite ${(x - 6)}^{2}$ as $(x - 6) (x - 6)$ .

$y = (x - 6) (x - 6) - 1$

Expand $(x - 6) (x - 6)$ using the FOIL Method.

Apply the distributive property.

$y = x (x - 6) - 6 (x - 6) - 1$

Apply the distributive property.

$y = x \cdot x + x \cdot - 6 - 6 (x - 6) - 1$

Apply the distributive property.

$y = x \cdot x + x \cdot - 6 - 6 x - 6 \cdot - 6 - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot - 6 - 6 x - 6 \cdot - 6 - 1$

Move $- 6$ to the left of $x$ .

$y = x^{2} - 6 \cdot x - 6 x - 6 \cdot - 6 - 1$

Multiply $- 6$ by $- 6$ .

$y = x^{2} - 6 x - 6 x + 36 - 1$

Subtract $6 x$ from $- 6 x$ .

$y = x^{2} - 12 x + 36 - 1$

Subtract $1$ from $36$ .

$y = x^{2} - 12 x + 35$

Step 3

Choose any value for $x$ that is in the domain to plug into the equation.

Step 4

Choose $0$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = 0^{2} - 12 \cdot 0 + 35$

Remove parentheses.

$y = {(0)}^{2} - 12 \cdot 0 + 35$

Simplify ${(0)}^{2} - 12 \cdot 0 + 35$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 - 12 \cdot 0 + 35$

Multiply $- 12$ by $0$ .

$y = 0 + 0 + 35$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 35$

Add $0$ and $35$ .

$y = 35$

Use the $x$ and $y$ values to form the ordered pair.

$(0, 35)$

Step 5

Choose $1$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = 1^{2} - 12 \cdot 1 + 35$

Remove parentheses.

$y = {(1)}^{2} - 12 \cdot 1 + 35$

Simplify ${(1)}^{2} - 12 \cdot 1 + 35$ .

Simplify each term.

One to any power is one.

$y = 1 - 12 \cdot 1 + 35$

Multiply $- 12$ by $1$ .

$y = 1 - 12 + 35$

Simplify by adding and subtracting.

Subtract $12$ from $1$ .

$y = - 11 + 35$

Add $- 11$ and $35$ .

$y = 24$

Use the $x$ and $y$ values to form the ordered pair.

$(1, 24)$

Step 6

Choose $2$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = 2^{2} - 12 \cdot 2 + 35$

Remove parentheses.

$y = {(2)}^{2} - 12 \cdot 2 + 35$

Simplify ${(2)}^{2} - 12 \cdot 2 + 35$ .

Simplify each term.

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

$y = 4 - 12 \cdot 2 + 35$

Multiply $- 12$ by $2$ .

$y = 4 - 24 + 35$

Simplify by adding and subtracting.

Subtract $24$ from $4$ .

$y = - 20 + 35$

Add $- 20$ and $35$ .

$y = 15$

Use the $x$ and $y$ values to form the ordered pair.

$(2, 15)$

Step 7

These are three possible solutions to the equation.

$(0, 35), (1, 24), (2, 15)$

Step 8

Find Three Ordered Pair Solutions h(x)=(x-6)^2-1

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