Find the Parabola Through (2,1) with Vertex (0,5) (0,5) , (2,1)

$(0, 5)$ , $(2, 1)$

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(0, 5)$ as the vertex $(h, k)$ and $(2, 1)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$1 = a {(2 - (0))}^{2} + 5$

Using $1 = a {(2 - (0))}^{2} + 5$ to solve for $a$ , $a = - 1$ .

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Rewrite the equation as $a {(2 - (0))}^{2} + 5 = 1$ .

$a {(2 - (0))}^{2} + 5 = 1$

Simplify each term.

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Subtract $0$ from $2$ .

$a \cdot 2^{2} + 5 = 1$

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

$a \cdot 4 + 5 = 1$

Move $4$ to the left of $a$ .

$4 a + 5 = 1$

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

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Subtract $5$ from both sides of the equation.

$4 a = 1 - 5$

Subtract $5$ from $1$ .

$4 a = - 4$

Divide each term by $4$ and simplify.

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Divide each term in $4 a = - 4$ by $4$ .

$\frac{4 a}{4} = \frac{- 4}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 a}{4} = \frac{- 4}{4}$

Divide $a$ by $1$ .

$a = \frac{- 4}{4}$

Divide $- 4$ by $4$ .

$a = - 1$

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(0, 5)$ and $a = - 1$ is $y = (- 1) {(x - (0))}^{2} + 5$ .

$y = (- 1) {(x - (0))}^{2} + 5$

Solve $y = (- 1) {(x - (0))}^{2} + 5$ for $y$ .

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Remove parentheses.

$y = (- 1) {(x - (0))}^{2} + 5$

Simplify each term.

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Subtract $0$ from $x$ .

$y = - 1 x^{2} + 5$

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

$y = - x^{2} + 5$

The standard form and vertex form are as follows.

Standard Form: $y = - x^{2} + 5$

Vertex Form: $y = (- 1) {(x - (0))}^{2} + 5$

Simplify the standard form.

Standard Form: $y = - x^{2} + 5$

Vertex Form: $y = - 1 x^{2} + 5$

Find the Parabola Through (2,1) with Vertex (0,5) (0,5) , (2,1)

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