Find the Parabola Through (2,6) with Vertex (1,3) (1,3) , (2,6)

$(1, 3)$ , $(2, 6)$

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 $(2, 6)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$6 = a {(2 - (1))}^{2} + 3$

Using $6 = a {(2 - (1))}^{2} + 3$ to solve for $a$ , $a = 3$ .

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Rewrite the equation as $a {(2 - (1))}^{2} + 3 = 6$ .

$a {(2 - (1))}^{2} + 3 = 6$

Simplify each term.

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Multiply $- 1$ by $1$ .

$a {(2 - 1)}^{2} + 3 = 6$

Subtract $1$ from $2$ .

$a \cdot 1^{2} + 3 = 6$

One to any power is one.

$a \cdot 1 + 3 = 6$

Multiply $a$ by $1$ .

$a + 3 = 6$

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

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Subtract $3$ from both sides of the equation.

$a = 6 - 3$

Subtract $3$ from $6$ .

$a = 3$

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

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

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

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Remove parentheses.

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

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

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Simplify each term.

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Multiply $- 1$ by $1$ .

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

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

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

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

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ .

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

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

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $- x$ .

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

Apply the distributive property.

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

Simplify.

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Multiply $- 2$ by $3$ .

$y = 3 x^{2} - 6 x + 3 \cdot 1 + 3$

Multiply $3$ by $1$ .

$y = 3 x^{2} - 6 x + 3 + 3$

Add $3$ and $3$ .

$y = 3 x^{2} - 6 x + 6$

The standard form and vertex form are as follows.

Standard Form: $y = 3 x^{2} - 6 x + 6$

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

Simplify the standard form.

Standard Form: $y = 3 x^{2} - 6 x + 6$

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

Find the Parabola Through (2,6) with Vertex (1,3) (1,3) , (2,6)

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