Algebra Examples

$(0, 4)$ , $(1, 3)$

Step 1

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(1, 3)$ as the vertex $(h, k)$ and $(0, 4)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$4 = a {(0 - (1))}^{2} + 3$

Step 2

Using $4 = a {(0 - (1))}^{2} + 3$ to solve for $a$ , $a = 1$ .

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Step 2.1

Rewrite the equation as $a {(0 - (1))}^{2} + 3 = 4$ .

$a {(0 - (1))}^{2} + 3 = 4$

Step 2.2

Simplify $a {(0 - (1))}^{2} + 3$ .

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Step 2.2.1

Subtract $1$ from $0$ .

$a {(- (1))}^{2} + 3 = 4$

Step 2.2.2

Simplify each term.

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Step 2.2.2.1

Multiply $- 1$ by $1$ .

$a {(- 1)}^{2} + 3 = 4$

Step 2.2.2.2

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

$a \cdot 1 + 3 = 4$

Step 2.2.2.3

Multiply $a$ by $1$ .

$a + 3 = 4$

Step 2.3

Move all terms not containing $a$ to the right side of the equation.

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Step 2.3.1

Subtract $3$ from both sides of the equation.

$a = 4 - 3$

Step 2.3.2

Subtract $3$ from $4$ .

$a = 1$

Step 3

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(1, 3)$ and $a = 1$ is $y = (1) {(x - (1))}^{2} + 3$ .

$y = (1) {(x - (1))}^{2} + 3$

Step 4

Solve $y = (1) {(x - (1))}^{2} + 3$ for $y$ .

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Step 4.1

Remove parentheses.

$y = (1) {(x - (1))}^{2} + 3$

Step 4.2

Multiply $1$ by ${(x - (1))}^{2}$ .

$y = 1 {(x - (1))}^{2} + 3$

Step 4.3

Remove parentheses.

$y = (1) {(x - (1))}^{2} + 3$

Step 4.4

Simplify $(1) {(x - (1))}^{2} + 3$ .

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Step 4.4.1

Simplify each term.

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Step 4.4.1.1

Multiply ${(x - (1))}^{2}$ by $1$ .

$y = {(x - (1))}^{2} + 3$

Step 4.4.1.2

Multiply $- 1$ by $1$ .

$y = {(x - 1)}^{2} + 3$

Step 4.4.1.3

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$y = (x - 1) (x - 1) + 3$

Step 4.4.1.4

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 4.4.1.4.1

Apply the distributive property.

$y = x (x - 1) - 1 (x - 1) + 3$

Step 4.4.1.4.2

Apply the distributive property.

$y = x \cdot x + x \cdot - 1 - 1 (x - 1) + 3$

Step 4.4.1.4.3

Apply the distributive property.

$y = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1 + 3$

Step 4.4.1.5

Simplify and combine like terms.

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Step 4.4.1.5.1

Simplify each term.

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Step 4.4.1.5.1.1

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1 + 3$

Step 4.4.1.5.1.2

Move $- 1$ to the left of $x$ .

$y = x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1 + 3$

Step 4.4.1.5.1.3

Rewrite $- 1 x$ as $- x$ .

$y = x^{2} - x - 1 x - 1 \cdot - 1 + 3$

Step 4.4.1.5.1.4

Rewrite $- 1 x$ as $- x$ .

$y = x^{2} - x - x - 1 \cdot - 1 + 3$

Step 4.4.1.5.1.5

Multiply $- 1$ by $- 1$ .

$y = x^{2} - x - x + 1 + 3$

Step 4.4.1.5.2

Subtract $x$ from $- x$ .

$y = x^{2} - 2 x + 1 + 3$

Step 4.4.2

Add $1$ and $3$ .

$y = x^{2} - 2 x + 4$

Step 5

The standard form and vertex form are as follows.

Standard Form: $y = x^{2} - 2 x + 4$

Vertex Form: $y = (1) {(x - (1))}^{2} + 3$

Step 6

Simplify the standard form.

Standard Form: $y = x^{2} - 2 x + 4$

Vertex Form: $y = {(x - 1)}^{2} + 3$

Step 7