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Algebra Examples

(0, 1)

$(0, 1)$ ,

(1, 0)

$(1, 0)$

Step 1

The general equation of a parabola with vertex

(h, k)

$(h, k)$ is

y = a {(x - h)}^{2} + k

$y = a {(x - h)}^{2} + k$ . In this case we have

(0, 1)

$(0, 1)$ as the vertex

(h, k)

$(h, k)$ and

(1, 0)

$(1, 0)$ is a point

(x, y)

$(x, y)$ on the parabola. To find

a

$a$ , substitute the two points in

y = a {(x - h)}^{2} + k

$y = a {(x - h)}^{2} + k$ .

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

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

Step 2

Using

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

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

a

$a$ ,

a = - 1

$a = - 1$ .

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Step 2.1

Rewrite the equation as

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

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

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

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

Step 2.2

Simplify each term.

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Step 2.2.1

Subtract

0

$0$ from

1

$1$ .

a \cdot 1^{2} + 1 = 0

$a \cdot 1^{2} + 1 = 0$

Step 2.2.2

One to any power is one.

a \cdot 1 + 1 = 0

$a \cdot 1 + 1 = 0$

Step 2.2.3

Multiply

a

$a$ by

1

$1$ .

a + 1 = 0

$a + 1 = 0$

a + 1 = 0

$a + 1 = 0$

Step 2.3

Subtract

1

$1$ from both sides of the equation.

a = - 1

$a = - 1$

a = - 1

$a = - 1$

Step 3

Using

y = a {(x - h)}^{2} + k

$y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex

(0, 1)

$(0, 1)$ and

a = - 1

$a = - 1$ is

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

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

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

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

Step 4

Solve

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

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

y

$y$ .

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Step 4.1

Remove parentheses.

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

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

Step 4.2

Multiply

- 1

$- 1$ by

{(x - (0))}^{2}

${(x - (0))}^{2}$ .

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

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

Step 4.3

Remove parentheses.

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

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

Step 4.4

Simplify each term.

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Step 4.4.1

Subtract

0

$0$ from

x

$x$ .

y = - 1 x^{2} + 1

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

Step 4.4.2

Rewrite

- 1 x^{2}

$- 1 x^{2}$ as

- x^{2}

$- x^{2}$ .

y = - x^{2} + 1

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

y = - x^{2} + 1

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

y = - x^{2} + 1

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

Step 5

The standard form and vertex form are as follows.

Standard Form:

y = - x^{2} + 1

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

Vertex Form:

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

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

Step 6

Simplify the standard form.

Standard Form:

y = - x^{2} + 1

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

Vertex Form:

y = - 1 x^{2} + 1

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

Step 7

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$