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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.

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

y=a(x−s)(x−t)

(s,0) and (t,0).

Because the points of a parabola with the same y-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).
From the graph, the following characteristics can be connected to the function rule.

$directionzerosaxis of symmetry :upward:(-4,0)&(-2,0):x=-3 →a>0→s=-4,t=-2→x=2-4+(-2) =-3 $

When quadratic functions are written in factored form y=a(x−s)(x−t), some of the parabola's characteristics are easy to identify.
### 1

To begin, identify the zeros, (s,0) and (t,0), from the function rule. Since the rule is y=-(x+1)(x−5),
### 2

Draw the axis of symmetry at x=2.
### 3

Next, determine the vertex. Since the axis of symmetry intersects the parabola at its vertex, the x-coordinate of the vertex is the same as the axis of symmetry. The corresponding y-coordinate can be found by substituting this x-value into the function rule and solving. Here, substitute x=2.
Thus, the vertex lies at (2,9). This point can be added to the graph.
### 4

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

$directionzerosaxis of symmetry :a>0⇒upward,a<0⇒downward:(s,0)and(t,0):x=2s+t $

It's possible to graph a quadratic function using its characteristics. Consider the function y=-(x+1)(x−5).
Identify and plot the zeros

$s=-1andt=5.$

Thus, the zeros are (-1,0) and (5,0). Next, plot these points on a coordinate plane.
Determine and draw the axis of symmetry

Points on the parabola with the same y-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 6 units apart.

Thus, the axis of symmetry is 3 units away from both of the zeros. Moving 3 units from either zero toward the other yields x=2. This can be verified algebraically using $x=2s+t .$Determine and plot the vertex

Draw the parabola

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(x−3), while Lewbowski's is modeled by L(x)=-2x2+6x. Note that x and y are both given in meters. The graph of L is shown.

Determine whose rocket went higher and whose went further.

Show 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 4.5 meters and the rocket traveled horizontally 3 meters from where it was launched at (0,0). Next, let's consider Allister's rocket. Recognizing that her function is given in factored form, we can write it asA(x)=-(x−0)(x−3).

It follows that the zeros are (0,0) and (3,0), which gives the horizontal distance of 3 meters. Thus, both rockets travel 3 meters from where they're launched. To find the maximum of A, we must first find the axis of symmetry. This is because the x-coordinate of the vertex is the same as the axis of symmetry.
A(x)=-x(x−3)

Substitute

x=1.5

A(1.5)=-1.5(1.5−3)

SubTerms

Subtract terms

A(1.5)=-1.5(-1.5)

MultNegNegOnePar

-a(-b)=a⋅b

A(1.5)=2.25

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