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Interpreting Quadratic Functions in Factored Form

Another way to represent quadratic functions is factored form, which expresses a quadratic function in terms of its factors, from which, the zeros of the parabola can be easily determined.
Rule

Factored Form of a Quadratic Function

Quadratic function rules can be expressed in factored form, sometimes referred to as intercept form.

y=a(xs)(xt)y=a(x-s)(x-t)

As is the case with standard form and vertex form, aa gives the direction of the parabola. When a>0,a>0, the parabola faces upward, and when a<0,a<0, it faces downward. Additionally, the zeros of the parabola lie at (s,0) and (t,0). (s,0) \text{ and } (t,0). Because the points of a parabola with the same yy-coordinate are equidistant from the axis of symmetry, the axis of symmetry lies halfway between the zeros. Consider the graph of f(x)=(x+4)(x+2).f(x)=(x+4)(x+2).

From the graph, the following characteristics can be connected to the function rule.

direction:upwarda>0zeros:(-4,0) & (-2,0)s=-4, t=-2axis of symmetry:x=-3x=-4+(-2)2=-3\begin{aligned} \textbf{direction} &: \text{upward} &&\rightarrow \hspace{-6pt} \quad a>0 \\ \\ \textbf{zeros}&: (\text{-} 4,0) \text{ \& } (\text{-} 2, 0 ) \hspace{-6pt} &&\rightarrow \hspace{-6pt} \quad s=\text{-} 4, \ t=\text{-} 2\\ \\ \textbf{axis of symmetry} &: x=\text{-} 3 &&\rightarrow \hspace{-6pt} \quad x=\frac{\text{-}4 + (\text{-}2)}{2}= \text{-} 3 \end{aligned}
Method

Graphing a Quadratic Function in Factored Form

When quadratic functions are written in factored form y=a(xs)(xt),y=a(x-s)(x-t), some of the parabola's characteristics are easy to identify. direction: a>0upward,    a<0downwardzeros:(s,0) and (t,0)axis of symmetry:x=s+t2\begin{aligned} \textbf{direction} &: \ a>0 \Rightarrow \text{upward,}\\ & \ \ \ \ a<0 \Rightarrow \text{downward}\\ \\ \textbf{zeros} &: (s,0) \text{ and } (t,0)\\ \\ \textbf{axis of symmetry} &: x=\frac{s+t}{2} \end{aligned} It's possible to graph a quadratic function using its characteristics. Consider the function y=-(x+1)(x5).y=\text{-}(x+1)(x-5).

1

Identify and plot the zeros

To begin, identify the zeros, (s,0)(s,0) and (t,0),(t,0), from the function rule. Since the rule is y=-(x+1)(x5),y=\text{-}(x+1)(x-5), s=-1andt=5. s=\text{-} 1 \quad \text{and} \quad t=5. Thus, the zeros are (-1,0)(\text{-} 1,0) and (5,0).(5,0). Next, plot these points on a coordinate plane.

2

Determine and draw the axis of symmetry

Points on the parabola with the same yy-coordinate are equidistant from the axis of symmetry. That means, the axis of symmetry is located halfway between the zeros. Notice that the zeros lie 66 units apart.

Thus, the axis of symmetry is 33 units away from both of the zeros. Moving 33 units from either zero toward the other yields x=2.x=2. This can be verified algebraically using x=s+t2.x=\frac{s+t}{2}. x=-1+52x=42=2 x=\dfrac{\text{-} 1+5}{2} \Rightarrow x=\dfrac{4}{2} = 2 Draw the axis of symmetry at x=2.x=2.

3

Determine and plot the vertex
Next, determine the vertex. Since the axis of symmetry intersects the parabola at its vertex, the xx-coordinate of the vertex is the same as the axis of symmetry. The corresponding yy-coordinate can be found by substituting this xx-value into the function rule and solving. Here, substitute x=2.x=2.
y=-(x+1)(x5)y=\text{-}(x+1)(x-5)
y=-(2+1)(25)y=\text{-}({\color{#0000FF}{2}}+1)({\color{#0000FF}{2}}-5)
y=-(3)(-3)y=\text{-}(3)(\text{-} 3)
y=9y=9
Thus, the vertex lies at (2,9).(2,9). This point can be added to the graph.

4

Draw the parabola

Considering the direction of the parabola, given by the aa-value in the function rule, the shape of the graph can be seen. Here, a=-1.a=\text{-} 1. Thus, the parabola opens downward. To draw the parabola, connect the points with a smooth curve.

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Exercise

Allister and Lebowski spend a Sunday afternoon launching bottle rockets in their backyard. For one round of launches, they collect the following data. The trajectory of Allister's rocket can be modeled by the function A(x)=-x(x3),A(x)=\text{-} x(x-3), while Lewbowski's is modeled by L(x)=-2x2+6x.L(x)=\text{-} 2x^2+6x. Note that xx and yy are both given in meters. The graph of LL is shown.

Determine whose rocket went higher and whose went further.

Show Solution
Solution

Here, we will compare two quadratic functions that model the trajectory of each rocket. To find the maximum height we'll need the vertex, and to find the horizontal distance we'll need the zeros. We'll begin with Lebowski's rocket.

From the graph, we can see that the maximum height is approximately 4.54.5 meters and the rocket traveled horizontally 33 meters from where it was launched at (0,0).(0,0). Next, let's consider Allister's rocket. Recognizing that her function is given in factored form, we can write it as A(x)=-(x0)(x3). A(x)=\text{-} (x-0)(x-3). It follows that the zeros are (0,0)(0,0) and (3,0)(3,0), which gives the horizontal distance of 33 meters. Thus, both rockets travel 33 meters from where they're launched. To find the maximum of A,A, we must first find the axis of symmetry. This is because the xx-coordinate of the vertex is the same as the axis of symmetry. x=s+t2=0+32=1.5 x=\dfrac{s+t}{2} =\frac{0+3}{2} = 1.5 Thus, x=1.5x=1.5 is the xx-coordinate of the vertex. We can substitute this into A(x)A(x) to find the corresponding yy-value, which will give the maximum height of Allister's rocket.
A(x)=-x(x3)A(x)=\text{-} x(x-3)
A(1.5)=-1.5(1.53)A({\color{#0000FF}{1.5}})=\text{-} {\color{#0000FF}{1.5}}({\color{#0000FF}{1.5}}-3)
A(1.5)=-1.5(-1.5)A(1.5)=\text{-} 1.5(\text{-} 1.5)
A(1.5)=2.25A(1.5)=2.25
Thus, Allister's rocket reached a maximum height of 2.252.25 meters. Since the maximum of Lebowski's rocket was 4.54.5 meters, it's clear his went higher. We can add the graph of A(x)A(x) to visually show our conclusions.
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