# What are the uses of ABCD parameters

### Stretching, compressing and shifting - the vertex shape

If you have quadratic functions of the form $$ f (x) = a * (x-d) ^ 2 + e $$, this is usually very useful.

You have already examined the parameters $$ a, d $$ and $$ e $$ individually. Now all 3 come together.

A functional equation of form

$$ f (x) = a * (x-d) ^ 2 + e $$ is called ** Vertex shape** the quadratic function.

### 1st example - reading and evaluating the parameter values

Given the equation of a quadratic function in vertex form, it reads:

$$ f (x) = 2 * (x-3) ^ 2 + 1 $$

You can read the following values for the parameters:

The values tell you that the normal parabola:

- is open to the top (because $$ a $$ is positive)
- is stretched (because $$ a> 1 $$)
- is shifted to the right (because $$ d $$ is positive)
- is shifted up (because $$ e $$ is positive)
- The parameters $$ d $$ and $$ e $$ give you the values for the vertex. The vertex is at $$ S (3 | 1) $$.

The coordinates of the vertex result from the values of the parameters $$ d $$ and $$ e $$.

### 2nd example - reading, evaluating and drawing the parabola

The given function is: $$ f (x) = 2 * (x + 4) ^ 2-3 $$.

### Reading and evaluating

- $$ a = + 2 $$: The normal parabola is open upwards and is stretched.
- $$ d = -4 $$: The normal parabola is shifted 4 units to the left
- $$ e = -3 $$: The normal parabola is shifted down three units.

The vertex is $$ S (-4 | -3) $$.

### Draw the parabola

Always start drawing the parabola by drawing in the vertex $$ S $$.

From the vertex you draw further points in the coordinate system.

With the normal parabola you go one unit to the right and then one unit up. However, since the normal parabola is inserted here with the factor $$ 2 $$, the $$ y $$ values are doubled. So you go one unit to the right and **two** Units up. Likewise, one step to the left and two steps up.

If you move two units to the right, you usually go up 4 units. But here you have to go up 8 units. You can find all other points according to the same pattern.

Connect the points to a parabola.

Always start drawing a parabola from the vertex.

You connect parabolas free-hand, not with a ruler.

You can only use the parabola template for a displaced normal parabola.

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### 3rd example - parabola opened at the bottom

The functional equation $$ f (x) = - 1/2 (x-2) ^ 2 + 1 $$ is given

### Reading and evaluating

- $$ a = -1 / 2 $$
- $$ d = + 2 $$
- $$ e = + 1 $$

The normal parabola is open downwards, it is compressed, shifted two units to the right and one unit upwards. The vertex is $$ S (2 | 1) $$.

### Draw the parabola

After you have drawn the vertex, you determine further points of the parabola. You go right or left as in the last example, but now you have to go down because the parabola is open downwards. The parameter $$ a $$ is the amount after $$ 1/2 $$, therefore the "normal" $$ y $$ values are halved. Go one unit to the right, then you have to go down half a unit $$ (1/2 * 1 = 1/2) $$. The same is true if you go one unit to the left.

If you go $$ 2 $$ units to the right or left, you have to go $$ 2 $$ units down $$ (1/2 * 4 = 2) $$.

The result is the following parabola:

### 4. Example - From graph to functional equation

Now it's the other way around. You want to get the function equation in the form $$ f (x) = a * (x-d) ^ 2 + e $$.

The following parabola is given:

First read the vertex: $$ S (2 | -3) $$.

That means the normal parabola was

- shifted 2 units to the right

$$ rarr $$ $$ d = + 2 $$ - moved down 3 units

$$ rarr $$ $$ e = -3 $$.

Then you determine the value of the parameter $$ a $$. You can see from the graph that $$ a $$ must be positive because the parabola is open upwards. Walk one unit to the right from the $$ S $$ vertex and then determine how many units you have to go up to hit the graph again.

$$ rarr $$ Here it is 1 unit.

If you move 2 units to the right, you then have to move 4 units upwards. This corresponds to the steps on the normal parabola, that is, this parabola is neither stretched nor compressed, so the value of the parameter $$ a = + 1 $$.

Plug all values into the vertex shape and you get: $$ f (x) = + 1 * (x-2) ^ 2-3 $$.

You can also omit the $$ + 1 $$:

$$ f (x) = (x-2) ^ 2-3 $$

### 5. Example - Difficult conditions

And another parable:

First read the vertex: $$ S (-1.5 | 0.5) $$. This means that $$ d = -1.5 $$ and $$ e = + 0.5 $$.

You immediately see that $$ a $$ must be negative, since the parabola is open downwards. To find the value for $$ a $$, you go one unit to the right from the apex and notice that the value you have to go down cannot be clearly read off. One could assume $$ a = -1 / 4 $$. To confirm this assumption, you go 2 units to the right and then only have to go one unit down $$ (- 1/4 * 4 = -1) $$. If you go 4 units to the right from the apex, you have to go 4 units down $$ (- 1/4 * 16 = -4) $$.

The value for the parameter $$ a $$ is really $$ - 1/4 $$.

Inserting into the vertex shape gives:

$$ f (x) = - 1/4 * (x + 1.5) ^ 2 + 0.5 $$

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### A note at the end

In order to find the function equation for a graph, the vertex and the value for the parameter $$ a $$ must be easy to read. This requires an exact coordinate system.

With this parabola you cannot or only imprecisely determine the function equation you are looking for.

The values of the parameters $$ a, d $$ and $$ e $$ have several decimal places. The functional equation for this parabola is:

$$ f (x) = 3/7 * (x + 1.283) ^ 2-2.085 $$

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