Forms of Quadratic Functions

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.

Fig. 2. Standard form.

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.

Fig. 3. Examples of standard form quadratic function on a graph.

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 \(x=3\).

Fig. 4. Factored form.

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.

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

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

Fig. 5. Vertex form.

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

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.

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 \(-3\).

Fig. 6. Example of conversion on a graph.

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:

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

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

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.

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.