If you look at these two pictures, what similarities can be seen between the bridge and the rainbow?
Both have the form of a parabola and can be described by quadratic functions. Of the highest point of parabolas is called apex. Do you now z. B. a quadratic function for the bridge, you can use the vertex shape determine its highest point.
Calculate vertex – basic knowledge
There are several ways to represent a quadratic function:
The prefactor may a never equal 0. the normal form is a special case of general shapein which the prefactor a = 1 is.
When you use which form depends on whether you want to find the zeros or the vertex of a quadratic function.
In memory of: The graph of quadratic functions is called a parabola and it opens either up or down.
For example, parabolas can look like this:
Figure 1: Examples of parabolas
Of the Vertex S describes either the highest (SG) or the lowest point (Sf) of a parabola.
Is the parabola after open at the top, is the vertex one rock bottomi.e. the lowest point.
Is the parabola after open belowthe vertex is a peaki.e. the highest point.
Vertex Shape – Definition
Any quadratic function can be mapped with the vertex form, also known as the vertex form.
The vertex form of a quadratic function is:
The apex of parabolas can be read directly from this form. This is then:
The letter a represents a prefactor that makes the parabola wider or narrower.
Determine or calculate the vertex
There are several ways to represent a quadratic function and find the vertex. Depending on which form of the quadratic function you have, the procedure differs.
Reading the vertex from the vertex shape
If the function is in vertex form, i.e. , you can read the vertex directly from it. This is then .
Pay attention to the signs. If in the brackets after the x a + is the x-value of your intersection (i.e) negative.
Conversion of normal form to vertex form
Given a function in normal form and you want to compute the vertex, you can convert that function to vertex form. You can do this with a square addition.
through the qsquare addition terms in which a variable occurs quadratically are transformed in such a way that the first or second binomial formula can be applied.
The binomial formulas help to raise sums and differences to the power.
First binomial formel:
Second binomial formula:
When converting, proceed as follows:
- First choose the appropriate binomial formula.
- Assign a and b the appropriate values of the function.
- Complete the binomial formula using square complement.
- Subtract the added term directly.
- Simplify the expression using the binomial formula.
Task 1
Write the following function in vertex form and calculate the vertex S:
solution
First, choose the binomial formula that fits the function. Since in the function with 6x added (positive sign), you use the first binomial formula.
Now you are looking for the right values for a and b. The value for a in the binomial formula xthere x just like a occurs in the square. To get the value for b To determine, just look at the part of the function that contains a single x and compare it to the expression 2ab.
So you divide the number that comes before the simple x by 2 and you get the value for b in the binomial formula. The value for a is always x!
There a = x applies, must b = 3 be.
The binomial formula for the function is as follows:
Since the term » » is still missing here, the square addition: You add the value for b as a square.
The binomial formula for the concrete example is in full form:
Since you can’t just add a number to an equation, you must use the same value subtract again:
In memory of: The +2 comes from the equation given above. It has been neglected so far, but now it has to be added back and not simply dropped.
Now you only use the binomial formula, so you simplify the first part of the term.
This is how you get the final crown shapefrom which you can read the vertex S:
In memory of: The vertex of a quadratic function is at .
The apex S reads:
Conversion General form to vertex form
It often happens that functions are not in the normal form, but in the general shape appear. They differ in the prefactor a.
You proceed in a similar way, you have to first step only the factor a exclude.
exercise 2
Write the following function in vertex form and calculate the vertex S.
solution
First, you factor out the prefactor. This is always in front of the x2, so it corresponds here 4.
This results in the following function:
Here it suffices to factor out the terms in which an x occurs. The 8 does not affect the binomial formula.
Here belongs the second binomial formula to the function term, because be deducted 3 times.
The value of a is always x. To determine b, you first only look at the part of the function in which a simple x occurs and divide it by 2.
The value in the binomial formula is for a = x and for .
So far, the binomial formula of this function is as follows:
along with the square complement The complete binomial formula is:
Next you subtract the added value, i.e. , again:
Make sure the parentheses include the . This is also multiplied by 4!
Now apply the binomial formula again and simplify the rest of the function:
You have now brought the function into vertex form. Now you can read the vertex:
The apex is then:
Conversion of factored form to vertex form
If there is a quadratic function in the factored form, you can read the zeros directly from it. They are: . The prefactor a is exactly the same as the vertex form or general form. It indicates how wide or narrow a parabola is.
As with vertex form, you need to pay special attention to the accidentals inside the parentheses. If the sign before x1 or x2 is negative, the zero is positive. However, if the sign is positive, the zero is negative.
The quadratic function
has the following zeros:
Figure 4: Parabola of f(x) with zeros x_1 and x_2 and vertex S
The apex S lies exactly in the middle between the two zeros x1 and x2. This applies to all parabolas.
This relationship between vertex and zeros helps convert the factored form to the vertex form. You derive the coordinates for the vertex S from the zeros. You proceed as follows:
- Reading the zeros from the factored form
- Determination of the x-coordinate i.e of the vertex with the mean of the zeros →
- Calculation of the y-coordinate e of the vertex insertion of i.e in f(x)
- Substituting into the vertex function
task 3
Convert the following function to vertex form and find the vertex.
solution
You read off the zeros of the function. They are as follows:
around the x coordinate i.e of the vertex S To determine, you calculate the mean between the two zeros x1 and x2.
This is possible because a parabola is axisymmetric to the y-axis. That is, it is reflected at the vertex. Accordingly, the vertex lies exactly in the middle between the two zeros.
The y coordinate e of the vertex S You calculate it by taking the x-coordinate i.e used in the function.
The vertex is .
You can now insert this into the vertex form with the vertex. The prefactor a you take over from the factored form:
The vertex form of this function as follows:
Calculate the vertex using the derivative
If you already sign up with Derivatives of Functions busy, you will find below another possibility to use the Svertex from the general form to calculate a function.
The vertex of a function in general form can be calculated by deriving the function.
In memory of: Derivatives indicate the slope of a function at the point x.
Viewed graphically, derivatives are to be understood as tangents, i.e. as linear functions, which in this case touch parabolas at a point.
If you draw such a tangent to the vertex of a parabola, you can see that the slope of this tangent is 0.
The function with the tangent is considered in the following example H at the apex.
Figure 5: Tangent at the vertex of a parabola
The slope at the vertex S of a parabola is always zero. This means that the derivative of a quadratic function is also zero at this point.
So you can find the x-value of the vertex by setting the derivative of the function equal to zero.
The function that you could already see as a parabola a little further up is considered:
The first step is to derive this function.
For derivations of this kind, the power rules are relevant. To delve deeper into the topic, take a look at the article!
Now the derivative is set equal to zero and solved for x:
This results in the x-value of the vertex S. For the y-value Du x = 1 into the original function, i.e. into , one:
The vertex is thus at .
Conversion of vertex form to general form
It can also sometimes be useful to convert a function that is in vertex form to general form.
On the general form you can z. B. the pq formulaor midnight formula to the Calculation of the zeros use.
For the conversion you do not need a quadratic complement, but multiply the term out.
task 4
Convert the following function to its general form.
solution
First you solve the binomial formula. This is the second binomial formula:
The second binomial formula is:
This results in the following term:
You put this resolved term into the function:
Now you multiply out. Start by resolving the parenthesis. Then you simplify the expression as much as possible:
Problems on vertex form
With the help of the following tasks you can now deepen your knowledge.
task 5
Convert the following function to vertex form and write the vertex S.
solution
1. Exclude prefactor
2. Select binomial formula
3. Determine a and b
4. square addition & direct subtraction
5. Apply & simplify binomial formula
The apex is therefore at .
Pay attention to the sign in the brackets when reading the vertex!
task 6
Convert the following function to vertex form and give the vertex.
solution
1. Reading the zeros
The zeros x1 and x2 ring:
2. Determination of the x-coordinate i.e of the vertex S with the mean of the zeros…