Find the Parabola Through (0,-15) with Vertex (1,-16) (1,-16) , (0,-15)

$(1, - 16)$ , $(0, - 15)$

Step 1

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(1, - 16)$ as the vertex $(h, k)$ and $(0, - 15)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$- 15 = a {(0 - (1))}^{2} - 16$

Step 2

Using $- 15 = a {(0 - (1))}^{2} - 16$ to solve for $a$ , $a = 1$ .

Rewrite the equation as $a {(0 - (1))}^{2} - 16 = - 15$ .

$a {(0 - (1))}^{2} - 16 = - 15$

Simplify $a {(0 - (1))}^{2} - 16$ .

Subtract $1$ from $0$ .

$a {(- (1))}^{2} - 16 = - 15$

Simplify each term.

Multiply $- 1$ by $1$ .

$a {(- 1)}^{2} - 16 = - 15$

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

$a \cdot 1 - 16 = - 15$

Multiply $a$ by $1$ .

$a - 16 = - 15$

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

Add $16$ to both sides of the equation.

$a = - 15 + 16$

Add $- 15$ and $16$ .

$a = 1$

Step 3

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(1, - 16)$ and $a = 1$ is $y = (1) {(x - (1))}^{2} - 16$ .

$y = (1) {(x - (1))}^{2} - 16$

Step 4

Solve $y = (1) {(x - (1))}^{2} - 16$ for $y$ .

Remove parentheses.

$y = (1) {(x - (1))}^{2} - 16$

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

$y = 1 {(x - (1))}^{2} - 16$

Remove parentheses.

$y = (1) {(x - (1))}^{2} - 16$

Simplify $(1) {(x - (1))}^{2} - 16$ .

Simplify each term.

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

$y = {(x - (1))}^{2} - 16$

Multiply $- 1$ by $1$ .

$y = {(x - 1)}^{2} - 16$

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

$y = (x - 1) (x - 1) - 16$

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

Apply the distributive property.

$y = x (x - 1) - 1 (x - 1) - 16$

Apply the distributive property.

$y = x \cdot x + x \cdot - 1 - 1 (x - 1) - 16$

Apply the distributive property.

$y = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1 - 16$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1 - 16$

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

$y = x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1 - 16$

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

$y = x^{2} - x - 1 x - 1 \cdot - 1 - 16$

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

$y = x^{2} - x - x - 1 \cdot - 1 - 16$

Multiply $- 1$ by $- 1$ .

$y = x^{2} - x - x + 1 - 16$

Subtract $x$ from $- x$ .

$y = x^{2} - 2 x + 1 - 16$

Subtract $16$ from $1$ .

$y = x^{2} - 2 x - 15$

Step 5

The standard form and vertex form are as follows.

Standard Form: $y = x^{2} - 2 x - 15$

Vertex Form: $y = (1) {(x - (1))}^{2} - 16$

Step 6

Simplify the standard form.

Standard Form: $y = x^{2} - 2 x - 15$

Vertex Form: $y = {(x - 1)}^{2} - 16$

Step 7

Find the Parabola Through (0,-15) with Vertex (1,-16) (1,-16) , (0,-15)

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