Find the Parabola Through (-1,6) with Vertex (0,4) (0,4) , (-1,6)

$(0, 4)$ , $(- 1, 6)$

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(0, 4)$ as the vertex $(h, k)$ and $(- 1, 6)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$6 = a {(- 1 - (0))}^{2} + 4$

Using $6 = a {(- 1 - (0))}^{2} + 4$ to solve for $a$ , $a = 2$ .

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Rewrite the equation as $a {(- 1 - (0))}^{2} + 4 = 6$ .

$a {(- 1 - (0))}^{2} + 4 = 6$

Simplify each term.

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Subtract $0$ from $- 1$ .

$a {(- 1)}^{2} + 4 = 6$

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

$a \cdot 1 + 4 = 6$

Multiply $a$ by $1$ .

$a + 4 = 6$

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

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Subtract $4$ from both sides of the equation.

$a = 6 - 4$

Subtract $4$ from $6$ .

$a = 2$

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(0, 4)$ and $a = 2$ is $y = (2) {(x - (0))}^{2} + 4$ .

$y = (2) {(x - (0))}^{2} + 4$

Solve $y = (2) {(x - (0))}^{2} + 4$ for $y$ .

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Remove parentheses.

$y = (2) {(x - (0))}^{2} + 4$

Subtract $0$ from $x$ .

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

The standard form and vertex form are as follows.

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

Vertex Form: $y = (2) {(x - (0))}^{2} + 4$

Simplify the standard form.

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

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

Find the Parabola Through (-1,6) with Vertex (0,4) (0,4) , (-1,6)

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