Arched paths are found for other activities involving projectiles, including shooting a cannonball and hitting a golf ball. In these scenarios, you can use Quadratic functions to learn how high the object will travel and where it will land.

In this explanation, we will explore the various forms of Quadratic functions, and see how to convert them from one to the other.

## What are the forms of quadratic functions?

There are three commonly used forms of quadratic functions.

**Standard or General Form**: \(y=ax^2+bx+c\)**Factored or Intercept Form**: \(y=a(bx+c)(dx+e)\)**Vertex Form**: \(y=a(x-h)^2+k\)

Each of these forms can be used to determine different information about the path of a projectile. Understanding the benefits of each form of a quadratic function will be useful for analyzing different situations that come your way.

## Standard form (general form) of a quadratic function

The graph of a quadratic function is a curve called a Parabola. All parabolas are symmetric with either a maximum (highest) or minimum (lowest) point. The point where a parabola meets its axis of symmetry is called the vertex. This vertex will either be the maximum or minimum point on the graph.

**Standard Form of a Quadratic Function**: \(f(x)=ax^2+bx+c\), where \(a, b\), and \(c\) are constants with \(a\neq 0\).

One benefit of standard form is that you can quickly identify the end behaviour and shape of the parabola by looking at the value of \(a\) in the function equation. This a-value is also referred to as the leading coefficient of the standard form equation. If the value of *a* is positive, the parabola opens upwards. If the value of \(a\) is negative, the parabola opens downwards.

Below is the graph of the quadratic function, \(f(x)=3x^2+2x-1\). Since this is a quadratic equation in standard form, we can see that \(a=3\). Notice that with a positive value of \(a\)*,* the parabola opens upwards.

Below is the graph of the quadratic function, \(f(x)=-3x^2+2x+1\). Since this is a quadratic equation in standard form, we can see that \(a=-3\). Notice that with a negative value of \(a\), the parabola opens downwards.

The standard form is helpful in

Finding the y-intercept. This can be done by setting \(x=0\).

Plugging into the quadratic formula by identifying the true values of \(a, b\), and \(c\).

Finding the axis of symmetry using \(x=\dfrac{-b}{2a}\).

## The factored form (intercept form) of a quadratic function

**Factored Form of a Quadratic Function**: \(f(x)=a(x-r_1)(x-r_2)\), where \(a\)* *is a constant and \(r_1\) and \(r_2\) are the roots of the function.

The factored form of a quadratic function, like the standard form, is useful in determining the end behaviour by analyzing the value of \(a\). As with standard form, the sign of *a* determines whether the parabola will open upwards or downwards.

The factored form has the added benefit of easily revealing the** **roots, or x-intercepts, of the function by application of the zero product property.

**Zero Product Property:** If \(a\times b=0\) then either \(a=0\) or \(b=0\).

For a quadratic function equation in the factored form \(f(x)=a(x-r_1)(x-r_2)\), we can apply the zero product property to find out when \(f(x)\) will be equal to zero. In other words, where \(x-r_1=0\) or \(x-r_2=0\) the graph will touch the x-axis.

Find the roots of the quadratic function \(f(x)=(2x+1)(x-4)\).

**Solution:**

When you are asked to find the roots of a function, you are being asked to find the x-values that result in \(f(x)=0\). In other words, you want to identify the x-intercepts.

Using the zero product property;

$$2x+1=0$$

or

$$x-4=0$$

Solve the first equation:

\[\begin{align} 2x+1&=0\\2x&=-1\\x&=-\dfrac{1}{2}\end{align}\]

Solving for the second equation:

\[\begin{align}x-4&=0\\x&=4\end{align}\]

Therefore, the roots of the function are \(x=-\dfrac{1}{2}\) and \(x=4\).

The graph of the parabola in factored form \(f(x)=-(x+2)(x-3)\) is facing downwards because \(a = -1\).

By applying the zero product property, we find that the roots are: \(x=-2\) and \(x=3\).

It is important to note that not all quadratic functions or equations have real roots. Some quadratics have imaginary numbers as their roots, and as a result, the factored form may not always be applicable.

## Vertex form of a quadratic function

**Vertex Form of a Quadratic Function**: \(f(x)=a(x-h)^2+k\), where \(a, h\)*, *and \(k\) are constants.

As indicated by its name, from vertex form, we can easily identify the vertex of the quadratic function using the values of \(h\) and \(k\). Also, as with standard and factored form, we can determine the end behaviour of the graph by looking at the a-value.

The quadratic function \(f(x)=-7(x-2)^2+16\) is in vertex form.

The value of \(a\) is \(-7\). Therefore, the graph will open downwards.

Recall that the vertex form of a quadratic equation is

$$f(x)=a(x-h)^2+k$$

and the equation given is

$$f(x)=-7(x-2)^2+16$$

By comparison, \(h\) is \(2\), while \(k\) is \(16\).

The vertex is \((2, 16)\) because \(h = 2\) and \(k = 16\).

The vertex is the point where the axis of symmetry meets the parabola. It is also the minimum point of a parabola that opens upwards or the maximum point of a parabola that opens downwards.

Consider the quadratic function \(f(x)=3(x-2)^2-1\) in the vertex form.

From the vertex form equation, \(a = 3\). Therefore, the graph opens upward.

Recall that the vertex form of a quadratic equation is

$$f(x)=a(x-h)^2+k$$

and the equation given is

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

By comparison, \(h\) is \(2\), while \(k\) is \(-1\).

Since \(h=2\) and \(k=-1\), the vertex is located at the point \((2,-1)\). This vertex is located on the axis of symmetry of the parabola. Therefore, the equation of the axis of symmetry for this quadratic function is \(x=2\). Notice, that the axis of symmetry is located at the x-value of the vertex.

## Converting between different forms of quadratic functions

Different scenarios may require you to solve for different key features of a parabola. It is useful to be able to convert the same quadratic function equation to different forms.

For instance, you may be asked to find the zeros, or x-intercepts, of a quadratic function equation given in the standard form. In order to efficiently find the zeros, we must first convert the equation to factored form.

### Converting a quadratic function from standard form to factored Form

Convert \(f(x)=2x^2+7x+3\) into factored form.

**Solution:**

To convert from the standard form into factored form, we need to factor the expression \(2x^2+7x+3\).

Let’s recall what Factored Form looks like this: \(f(x)=a(x-r_1)(x-r_2)\).

In order to factor the expression, we can factor the expression by grouping.

To do this, find the factors of the product of the values of \(a\) and \(c\) that also sum up to make \(b\). In this case, \(6\) is the product of \(a\) and \(c\), and \(b=7\). We can list the factors of \(6\) and their sums as follows:

Factors of \(6\);

- \(1\) and \(6\) : \(1+6=7\)
- \(2\) and \(3\) : \(2+3=5\)

The two values whose product is \(6\) and sum up to \(7\) are \(1\) and \(6\). We can now split the middle term and rewrite the expression as follows:

$$2x^2+7x+3=(2x^2+6x)+(x+3)$$

Now we can factor out the GCF of each group. In this case, \(2x\) can be factored out of the first two terms and \(1\) can be factored out of the last two terms. Therefore, we can factor the entire expression by applying the distributive property.

$$2x(x+3)+1(x+3)$$

$$(2x+1)(x+3)$$

Therefore, our resulting equation in factored form is \(f(x)=(2x+1)(x+3)\).

Now we can proceed to find the zeros, roots, or x-intercepts by setting the function equation equal to zero and applying the zero product property.

$$(2x+1)(x+3)=0$$

$$2x+1=0$$

$$2x=-1$$

$$x=-\dfrac{1}{2}$$

or

$$x+3=0$$

$$x=-3$$

Therefore, the zeros of the function \(f(x)=2x^2+7x+3\) are \(-\dfrac{1}{2}\) and \(-3\).

### Converting a quadratic function from standard form to vertex form

Instead of solving for the zeros of a quadratic function, we could instead be asked for the vertex. For instance, we could be asked to find the vertex of a quadratic function or equation.

To find the vertex, it would be helpful to convert the standard form equation into vertex form.

Remember, the vertex form of the quadratic function equation is \(f(x)=a(x-h)^2+k\).

To switch from standard form to vertex form, we can use a strategy called** completing the square.** Basically, we are using algebraic reasoning to create a trinomial that can be factored into a perfect square.

**Perfect Square Trinomial**: an expression that is obtained by squaring a binomial equation. It is in the form \(a^2+2ab+b^2=(a+b)^2\).

Simply put, we need to strategically choose a constant to add to the equation that allows up to factor the expression as a perfect square. This will create the \((x-h)^2\) part of the vertex form equation.

Convert the quadratic function \(f(x)=-3x^2-6x-9\) into vertex form.

**Solution:**

Step 1:

If we have a leading coefficient other than one, we can factor that value outside of the trinomial as a common factor. Recall that the leading coefficient is the number in front of \(x^2\). In this case, the leading coefficient is \(-3\).

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

Step 2:

We need to determine which value to add to the equation that will create a perfect square trinomial on one side. This value will always be \(\left(\dfrac{b}{2}\right)^2\). In our resulting trinomial, \(b = 2\). Therefore:

$$\left(\dfrac{2}{2}\right)^2=1^2=1$$

Now we can add this value as a constant within our trinomial. You may be thinking, "how are we allowed to choose a number to add to the trinomial?" We can only add the value if we also subtract it! That way, we are effectively adding \(0\) to the trinomial. The result will look like this:

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

Notice that by so doing we have obtained a perfect square trinomial (thus, the strategy name “completing the square”). Now we have created a perfect square trinomial as the first three terms in the bracket which we can factor into the square of a binomial.

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

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

Distributing the \(-3\) results in the following:

$$y=-3(x+1)^2-6$$

Recall that the vertex form of a quadratic equation is expressed as

$$f(x)=a(x-h)^2+k$$

and you have

$$y=-3(x+1)^2-6$$

hence, \(h\) is \(-1\), while \(k\) is \(-6\).

We now have our quadratic equation in vertex form. In this form, we see that the vertex, \((h,k)\) is \((-1,-6)\).

### Converting a quadratic function from factored form to standard form

Converting a quadratic function equation from the factored form into standard form involves multiplying the factors. You can do this by applying the distributive property, sometimes referred to as the FOIL method.

Convert the quadratic function \(f(x)=(3x-2)(-x+7)\) into standard form.

**Solution:**

Using double distribution, or FOIL, we multiply the factors \((3x-2)\) and \((-x+7)\) together. Thus:

$$f(x)=(3x)(-x)+(3x)(7)+(-2)(-x)+(-2)(7)$$

$$f(x)=-3x^2+21x+2x-14$$

$$f(x)=-3x^2+23x-14$$

We now have the equation rewritten in standard form. From here, we can identify the axis of symmetry and the y-intercept.

### Converting a quadratic function from vertex form to standard form

Finally, there may also be situations where you need to convert a quadratic function equation from vertex form into standard form.

Convert the equation \(f(x)=2(x+7)^2-10\) into standard form.

**Solution:**

We shall expand the expression \((x+7)^2\), again using double distribution to multiply. Then, distribute the a-value throughout the resulting trinomial. Finally, combine like terms.

\[\begin{align}f(x)&=2(x+7)^2-10=\\&=2(x+7)(x+7)-10=\\&=2(x^2+14x+49)-10=\\&=2x^2+28x+98-10=\\&=2x^2+28x+88\end{align}\]

We now have the equation rewritten in standard form. Once again, we can identify the axis of symmetry and y-intercept.

## Forms of Quadratic Functions - Key takeaways

- The graph of a quadratic function is a curve called a parabola. Parabolas have several key features of interest including end behavior, zeros, an axis of symmetry, a y-intercept, and a vertex.
- The standard form of a quadratic function equation is \(f(x)=ax^2+bx+c\), where \(a, b\), and \(c\) are constants with \(a\neq0\).
- Standard form allows us to easily identify: end behaviour, the axis of symmetry, and y-intercept.
- The factored form of a quadratic function is \(f(x)=a(x-r_1)(x-r_2)\).
- Factored form allows us to easily identify: end behaviour, and zeros.
- The vertex form of a quadratic function is \(f(x)=a(x-h)^2+k\), where \(a, h\), and \(k\) are constants with \(a\neq 0\).
- Vertex form allows us to easily identify: end behaviour, and vertex.
- We can use polynomial multiplication and factoring principles to convert between these different forms.

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##### Frequently Asked Questions about Forms of Quadratic Functions

What are forms of quadratic functions?

There are three forms of quadratic functions such as the standard or general form, factored or intercept form, and the vertex form.

What is the vertex form of a quadratic function?

The vertex form of a quadratic function is expressed as: y=a(x-h)^{2}+k, where *a, h, *and *k* are constants.

What is the factored form of a quadratic function?

The factored form of a quadratic function is expressed as: y=a(x-r_{1})(x-r_{2}), where *a *is a constant and r_{1} and r_{2} are the roots of the function.

What is the standard form of a quadratic function?

The standard form of a quadratic function is expressed as: y=ax^{2}+bx+c , where a, b, and c are constants with a≠0.

How to find the factored form of a quadratic function?

The factored form of a quadratic equation is found by expressing the equation in the form f(x)=a(x-r_{1})(x-r_{2}), where *a *is a constant and r_{1} and r_{2} are the roots of the function.

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