Find the Parabola Through (0,-3) with Vertex (1,1) (1,1) , (0,-3)

$(1, 1)$ , $(0, - 3)$

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(1, 1)$ as the vertex $(h, k)$ and $(0, - 3)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$- 3 = a {(0 - (1))}^{2} + 1$

Using $- 3 = a {(0 - (1))}^{2} + 1$ to solve for $a$ , $a = - 4$ .

Rewrite the equation as $a {(0 - (1))}^{2} + 1 = - 3$ .

$a {(0 - (1))}^{2} + 1 = - 3$

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

Subtract $1$ from $0$ .

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

Simplify each term.

Multiply $- 1$ by $1$ .

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

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

$a \cdot 1 + 1 = - 3$

Multiply $a$ by $1$ .

$a + 1 = - 3$

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

Subtract $1$ from both sides of the equation.

$a = - 3 - 1$

Subtract $1$ from $- 3$ .

$a = - 4$

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(1, 1)$ and $a = - 4$ is $y = (- 4) {(x - (1))}^{2} + 1$ .

$y = (- 4) {(x - (1))}^{2} + 1$

Solve $y = (- 4) {(x - (1))}^{2} + 1$ for $y$ .

Remove parentheses.

$y = (- 4) {(x - (1))}^{2} + 1$

Multiply $- 4$ by ${(x - (1))}^{2}$ .

$y = - 4 {(x - (1))}^{2} + 1$

Remove parentheses.

$y = (- 4) {(x - (1))}^{2} + 1$

Simplify $(- 4) {(x - (1))}^{2} + 1$ .

Simplify each term.

Multiply $- 1$ by $1$ .

$y = - 4 {(x - 1)}^{2} + 1$

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

$y = - 4 ((x - 1) (x - 1)) + 1$

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

Apply the distributive property.

$y = - 4 (x (x - 1) - 1 (x - 1)) + 1$

Apply the distributive property.

$y = - 4 (x \cdot x + x \cdot - 1 - 1 (x - 1)) + 1$

Apply the distributive property.

$y = - 4 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) + 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = - 4 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) + 1$

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

$y = - 4 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) + 1$

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

$y = - 4 (x^{2} - x - 1 x - 1 \cdot - 1) + 1$

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

$y = - 4 (x^{2} - x - x - 1 \cdot - 1) + 1$

Multiply $- 1$ by $- 1$ .

$y = - 4 (x^{2} - x - x + 1) + 1$

Subtract $x$ from $- x$ .

$y = - 4 (x^{2} - 2 x + 1) + 1$

Apply the distributive property.

$y = - 4 x^{2} - 4 (- 2 x) - 4 \cdot 1 + 1$

Simplify.

Multiply $- 2$ by $- 4$ .

$y = - 4 x^{2} + 8 x - 4 \cdot 1 + 1$

Multiply $- 4$ by $1$ .

$y = - 4 x^{2} + 8 x - 4 + 1$

Add $- 4$ and $1$ .

$y = - 4 x^{2} + 8 x - 3$

The standard form and vertex form are as follows.

Standard Form: $y = - 4 x^{2} + 8 x - 3$

Vertex Form: $y = (- 4) {(x - (1))}^{2} + 1$

Simplify the standard form.

Standard Form: $y = - 4 x^{2} + 8 x - 3$

Vertex Form: $y = - 4 {(x - 1)}^{2} + 1$

Find the Parabola Through (0,-3) with Vertex (1,1) (1,1) , (0,-3)

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