What are the Standard Form and Vertex Form? [pptime time="0:01"](0:01) [/pptime]

Quadratic equations can be written in two different forms: Standard Form (also known as the General Form ): $$ y = ax^2 + bx + c $$ Vertex Form : $$ y = a(x\ - \ h)^2 + k $$ In both forms, $ a \ne 0 $. These are just two ways of expressing the same quadratic relationship: Use the standard form $ y = ax^2 + bx + c $ when you're solving or simplifying algebraically. Use the vertex form $ y = a(x\ - \ h)^2 + k $ when you want to find the vertex easily. So, you can switch between the two forms depending on what the question is asking.

How to convert Vertex Form to Standard Form? [pptime time="0:17"](0:17) [/pptime]

To convert from vertex form to standard form, follow these steps: $$ y = 2(x\ - \ 3)^2\ - \ 5 $$ Step 1: Square the Binomial Applying the formula: $$ (a\ - \ b)^2 = a^2\ - \ 2ab + b^2 $$ $$ (x\ - \ 3)^2 = x^2\ - \ 6x + 9 $$ Step 2: Expand the brackets Multiply each term inside the bracket by 2: $$ \begin{aligned} y &= 2(x^2\ - \ 6x + 9) \ - \ 5 \\[0.8em] &= 2x^2\ - \ 12x + 18 \ - \ 5 \\[0.8em] \end{aligned} $$ Step 3: Simplify Combine like terms: + 18 - 5 becomes + 13 So the standard form is: $$ y = 2x^2 \ - \ 12x + 13 $$

How to convert Standard Form to Vertex Form? [pptime time="1:12"](1:12) [/pptime]

To convert from standard form to vertex form, follow these steps: $$ y = 3x^2 + 18x + 25 $$ Step 1: Factor out the coefficient of $x^2$ The coefficient of $x^2$ is 3, so we factor out 3 from the first two terms: the $x^2$ and $x$ terms. $$ y = 3(x^2 + 6x) + 25 $$ Step 2: Complete the square Take half of 6 (which is 3), square it (which is 9), then add and subtract it inside the bracket: $$ y = 3(x^2 + 6x + 9 \ - \ 9) + 25 $$ Group the first 3 terms to form a perfect square: $$ y = 3\left[(x^2 + 6x + 9)\ - \ 9\right] + 25 $$ Now write the trinomial as a square: $$ y = 3\left[(x + 3)^2\ - \ 9\right] + 25 $$ Step 3: Simplify Distribute the 3 (and don't forget the constant!) $$ y= 3(x + 3)^2\ - \ 27 + 25 $$ Combine like terms: - 27 +25 becomes - 2 So the vertex form is: $$ y = 3(x + 3)^2 \ - \ 2 $$

Converting General Form to Vertex Form (Example) [pptime time="2:21"](2:21) [/pptime]

To convert from general form to vertex form, follow these steps: $$ y = -2x^2 + 8x \ - \ 5 $$ Step 1: Factor out the coefficient of $x^2$ The coefficient of $x^2$ is $-2$, so we factor out $-2$ from the first two terms: the $x^2$ and $x$ terms. $$ y = -2(x^2\ - \ 4x)\ - \ 5 $$ Step 2: Complete the square Take half of $-4$ (which is $-2$), square it (which is $4$), then add and subtract it inside the bracket: $$ y = -2(x^2\ - \ 4x + 4\ - \ 4)\ - \ 5 $$ Group the first 3 terms to form a perfect square: $$ y = -2\left[(x^2\ - \ 4x + 4) \ - \ 4\right]\ - \ 5 $$ Now write the trinomial as a square: $$ y = -2\left[(x\ - \ 2)^2\ - \ 4\right]\ - \ 5 $$ Step 3: Simplify Distribute the $-2$ and simplify: $$ y = -2(x\ - \ 2)^2 + 8 \ - \ 5 $$ Combine like terms: $+8 - 5$ becomes $+3$ So the vertex form is: $$ y = -2(x\ - \ 2)^2 + 3 $$

Converting Quadratics: Standard Form and Vertex Form

🎬 Video: Converting Quadratics Step by Step

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What are the Standard Form and Vertex Form? (0:01)

Quadratic equations can be written in two different forms:

Standard Form (also known as the General Form):

$$
y = ax^2 + bx + c
$$

Vertex Form:

$$
y = a(x\ – \ h)^2 + k
$$

In both forms, $ a \ne 0 $.

These are just two ways of expressing the same quadratic relationship:

Use the standard form $ y = ax^2 + bx + c $ when you’re solving or simplifying algebraically.
Use the vertex form $ y = a(x\ – \ h)^2 + k $ when you want to find the vertex easily.

So, you can switch between the two forms depending on what the question is asking.

How to convert Vertex Form to Standard Form? (0:17)

To convert from vertex form to standard form, follow these steps:

$$
y = 2(x\ – \ 3)^2\ – \ 5
$$

Step 1: Square the Binomial
Applying the formula:

$$
(a\ – \ b)^2 = a^2\ – \ 2ab + b^2
$$

$$
(x\ – \ 3)^2 = x^2\ – \ 6x + 9
$$

Step 2: Expand the brackets
Multiply each term inside the bracket by 2:

$$
\begin{aligned}
y &= 2(x^2\ – \ 6x + 9) \ – \ 5 \\[0.8em]
&= 2x^2\ – \ 12x + 18 \ – \ 5 \\[0.8em]
\end{aligned}
$$

Step 3: Simplify
Combine like terms: + 18 – 5 becomes + 13
So the standard form is:

$$
y = 2x^2 \ – \ 12x + 13
$$

How to convert Standard Form to Vertex Form? (1:12)

To convert from standard form to vertex form, follow these steps:

$$
y = 3x^2 + 18x + 25
$$

Step 1: Factor out the coefficient of $x^2$
The coefficient of $x^2$ is 3, so we factor out 3 from the first two terms: the $x^2$ and $x$ terms.

$$
y = 3(x^2 + 6x) + 25
$$

Step 2: Complete the square
- Take half of 6 (which is 3), square it (which is 9), then add and subtract it inside the bracket:

$$
y = 3(x^2 + 6x + 9 \ – \ 9) + 25
$$

- Group the first 3 terms to form a perfect square:

$$
y = 3\left[(x^2 + 6x + 9)\ – \ 9\right] + 25
$$

- Now write the trinomial as a square:

$$
y = 3\left[(x + 3)^2\ – \ 9\right] + 25
$$

Step 3: Simplify
- Distribute the 3 (and don’t forget the constant!)

$$
y= 3(x + 3)^2\ – \ 27 + 25
$$

- Combine like terms: – 27 +25 becomes – 2
  So the vertex form is:

$$
y = 3(x + 3)^2 \ – \ 2
$$

Converting General Form to Vertex Form (Example) (2:21)

To convert from general form to vertex form, follow these steps:

$$
y = -2x^2 + 8x \ – \ 5
$$

Step 1: Factor out the coefficient of $x^2$
The coefficient of $x^2$ is $-2$, so we factor out $-2$ from the first two terms: the $x^2$ and $x$ terms.

$$
y = -2(x^2\ – \ 4x)\ – \ 5
$$

Step 2: Complete the square
- Take half of $-4$ (which is $-2$), square it (which is $4$), then add and subtract it inside the bracket:

$$
y = -2(x^2\ – \ 4x + 4\ – \ 4)\ – \ 5
$$

- Group the first 3 terms to form a perfect square:

$$
y = -2\left[(x^2\ – \ 4x + 4) \ – \ 4\right]\ – \ 5
$$

- Now write the trinomial as a square:

$$
y = -2\left[(x\ – \ 2)^2\ – \ 4\right]\ – \ 5
$$

Step 3: Simplify
Distribute the $-2$ and simplify:

$$
y = -2(x\ – \ 2)^2 + 8 \ – \ 5
$$

Combine like terms: $+8 – 5$ becomes $+3$
So the vertex form is:

$$
y = -2(x\ – \ 2)^2 + 3
$$

📂 Flashcards: Convert Between Standard and Vertex Form

Converting quadratic equations between standard form y = ax² + bx + c and vertex form y = a(x - h)² + k. — 1) Standard Form and Vertex Form

Converting a quadratic equation from vertex form, y = 2(x - 3)² - 5, to standard form by expanding brackets and simplifying to y = 2x² - 12x + 13. — 2) Converting Vertex Form to Standard Form

Conversion of a quadratic equation from standard form y = 3x² + 18x + 25 to vertex form by completing the square, resulting in y = 3(x + 3)² - 2. — 3) Converting Standard Form to Vertex Form

Converting the quadratic equation y = -2x² + 8x - 5 from general to vertex form using completing the square method, resulting in y = -2(x - 2)² + 3. — 4) Converting General Form to Vertex Form

🍪 Quiz: Test Your Skills on Converting Quadratic Forms

Converting Quadratics: General and Vertex Form

1 / 6

Q: Convert the quadratic equation $y = 3(x + 2)^2 - 4$ into general form.

$y = 3x^2 + 12x + 4$

$y = 3x^2 - 6x - 4$

$y = 3x^2 + 12x + 8$

$y = 3x^2 + 6x - 8$

2 / 6

Q: Convert the quadratic equation $y = 4(x - 1)^2 + 2$ into general form.

$y = 4x^2 - 4x + 6$

$y = 4x^2 - 8x + 6$

$y = 4x^2 - 8x + 10$

$y = 4x^2 - 4x + 10$

3 / 6

Q: Convert the quadratic equation $y = x^2 + 4x + 5$ into vertex form.

$y = (x - 2)^2 - 1$

$y = (x - 2)^2 + 1$

$y = (x + 2)^2 - 1$

$y = (x + 2)^2 + 1$

4 / 6

Q: Convert the quadratic equation $y = x^2 - 6x + 8$ into vertex form.

$y = (x - 6)^2 + 1$

$y = (x + 6)^2 - 1$

$y = (x - 3)^2 - 1$

$y = (x + 3)^2 + 1$

5 / 6

Q: Convert the quadratic equation $y = x^2 - 8x + 15$ into vertex form.

$y = (x - 8)^2 - 1$

$y = (x + 4)^2 + 1$

$y = (x - 4)^2 - 1$

$y = (x + 2)^2 - 1$

6 / 6

Q: Convert the quadratic equation $y = 3x^2 - 12x + 5$ into vertex form.

$y = 3(x + 2)^2 - 7$

$y = 3(x - 2)^2 + 7$

$y = 3(x + 2)^2 + 7$

$y = 3(x - 2)^2 - 7$

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