Wize High School Algebra I Textbook (Common Core) > Quadratic Functions

Factored Form of a Quadratic Equation

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Factored Form of a Quadratic Equation
Recall
The standard/ general form of a quadratic equation is y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c.
We can solve problems involving quadratic relations by sketching its graph
We can sketch its graph by creating a table of values
Graphing with a table of values can be very tedious and inaccurate without the help of technology.

Factored Form
Instead, we turn to a different form-- the factored form of a quadratic equation is y=a(x−r)(x−s)y=a(x-r)(x-s)y=a(x−r)(x−s).

*(x−r)(x-r)(x−r) and (x−s)(x-s)(x−s) are called the factors of this quadratic equation.


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How do we graph from the factored form?
Direction of the opening
The aaa value tells us the           direction of the opening          
If aaa is positive, the parabola      opens up     
If aaa is negative, the parabola      opens down     
Zeros and roots
The x-intercepts/ roots/ zeros of a quadratic graph in factored form are           x=r   and   x=s          
If the factored form of the quadratic equation is y=a(x−r)(x−s)y=a(x-r)(x-s)y=a(x−r)(x−s), where rrr and sss are different, then                          there are 2 different zeros (r,0) and (s,0)                         
If the factored form of the quadratic equation is y=a(x−r)2y=a(x-r)^2y=a(x−r)2, then                     there is 1 single zero (r,0)                    
If we cannot find the factored form of the quadratic equation y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c, then                there are no zeros               
Vertex
To find the coordinates of the vertex (h,k)(h,k)(h,k):
The x-coordinate of the vertex is directly in between the two zeros -- h=r+s2h=\dfrac{r+s}{2}h=2r+s​
Substitute this x-coordinate into the quadratic equation to dtermine kkk, the y-coordinate of the vertex

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Example: Factored Form of a Quadratic Equation
Given the factored form y=(x+2)(x−5)y=(x+2)(x-5)y=(x+2)(x−5),

a) determine the direction of opening of the quadratic graph.

The factored form is y=1(x+2)(x−5)y=\bct1\left(x+2\right)\left(x-5\right)y=1(x+2)(x−5), so a=1a=1a=1, meaning that the parabola will be opening up\boxed{\text{opening up}}opening up​.

b) determine the zeros.

The zeros/ x-intercepts/ roots is when the y value is 0: 0=(x+2)(x−5)0=\left(x+2\right)\left(x-5\right)0=(x+2)(x−5)

Notice that if either of the brackets equal 0, then the y value will be 0:
x+2=0orx−5=0x=−2orx=5\begin{array}{ccc}
x+2=0&\text{or}&x-5=0\\
x=-2&\text{or}&x=5
\end{array}x+2=0x=−2​oror​x−5=0x=5​
So, the zeros are x=−2x=-2x=−2 and x=5x=5x=5, which correspond to the points (−2,0)   and   (5,0)\boxed{(-2, 0)~~~\text{and}~~~(5,0)}(−2,0)   and   (5,0)​.

*Notice that these are the rrr and sss values in the factored form y=a(x−r)(x−s)y=a(x-r)(x-s)y=a(x−r)(x−s).

c) determine the y-intercept.

y-intercept is where the graph meets the y-axis, meaning when x=0x=0x=0:
y=(0+2)(0−5)=(2)(−5)=−10\begin{aligned}
y&=(0+2)(0-5)\\
&=(2)(-5)\\
&=-10
\end{aligned}y​=(0+2)(0−5)=(2)(−5)=−10​
So, the y-intercept is y=10y=10y=10, which corresponds to the point (0,−10)\boxed{(0,-10)}(0,−10)​.

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d) determine the axis of symmetry.

Since the zeros have the same y-values, they are reflections of one another along the axis of symmetry, meaning that the axis of symmetry should be right in the middle between the 2 zeros:
−2+52=32\dfrac{-2+5}{2}=\dfrac{3}{2}2−2+5​=23​
So, the axis of symmetry is x=32\boxed{x=\dfrac{3}{2}}x=23​​

e) determine the coordinates of the vertex.

The x-coordinate of the vertex is the same as the axis of symmetry: x=32x=\dfrac{3}{2}x=23​.

To find the y-coordinate, we substitute x=32x=\dfrac{3}{2}x=23​ back into the quadratic equation y=(x+2)(x−5)y=(x+2)(x-5)y=(x+2)(x−5):
y=(x+2)(x−5)=(32+2)(32−5)=(72)(−72)=−494      (or 12.25)\begin{aligned}
y&=\left(\colorbox{yellow}{$x$}+2\right)\left(\colorbox{yellow}{$x$}-5\right)\\
&=\left(\colorbox{yellow}{$\dfrac{3}{2}$}+2\right)\left(\colorbox{yellow}{$\dfrac{3}{2}$}-5\right)\\
&=\left(\dfrac{7}{2}\right)\left(-
\dfrac{7}{2}\right)\\
&=-\dfrac{49}{4}~~~~~~(\text{or }12.25)
\end{aligned}y​=(x​+2)(x​−5)=(23​​+2)(23​​−5)=(27​)(−27​)=−449​      (or 12.25)​

So, the vertex is (32,−494)\boxed{\left(\dfrac{3}{2},-\dfrac{49}{4}\right)}(23​,−449​)​

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f) determine the standard form of the quadratic equation.

To go from the factored form to the standard form of a quadratic equation, we need to expand and simplify the equation:
     (x+2)(x−5)=(x)(x)+(x)(−5)+(2)(x)+(2)(−5)=     x2       −5x          +2x       −10=     x2                −3x                 −10\begin{aligned}
&~~~~~(\bct{x}\bcth{+2})(x-5)\\
&=\bct{(x)}(x)+\bct{(x)}(-5)+\bcth{(2)}(x)+\bcth{(2)}(-5)\\
&=~~~~~x^2~~~~~~~-5x~~~~~~~~~~+2x~~~~~~~-10\\
&=~~~~~x^2~~~~~~~~~~~~~~~~-3x~~~~~~~~~~~~~~~~~-10
\end{aligned}​     (x+2)(x−5)=(x)(x)+(x)(−5)+(2)(x)+(2)(−5)=     x2       −5x          +2x       −10=     x2                −3x                 −10​

So, the standard form is x2−3x−10\boxed{x^2-3x-10}x2−3x−10​

g) sketch the graph (without using a table of values or technology).

Practice: Factored Form of a Quadratic Equation
Determine the zeros of the following quadratic equations.

Select all of the zeros that apply.

Practice: Factored Form of a Quadratic Equation
For each of the following quadratic equations,
i) determine the zeros
ii) determine the vertex
iii) determine the y-intercept
iv) graph the quadratic equation.

y=−3x(x+2)y=-3x\left(x+2\right)y=−3x(x+2)

i) Enter the ZERO (x-intercept) with the SMALLER x value first, as a point in the form

(x,y)

i) Enter the ZERO (x-intercept) with the LARGER x-value, as a point in the form

(x,y)

ii) Enter the VERTEX as a point in the form

(x,y)

iii) Enter the Y-INTERCEPT as a point in the form

(x,y)

iv) Check your graph with the solution

I don't know

Practice: Factored Form of a Quadratic Equation
You are walking through a forest and you see a bridge in front of you. The height of the bridge above the ground (in meters) can be modelled by the equation h=−0.5(x−1)(x−25)h=-0.5(x-1)(x-25)h=−0.5(x−1)(x−25), where xxx is the distance (in meters) from you to the start of the bridge.

a) How far away is the start of the bridge from you?

b) What is the horizontal length of this bridge? (Meaning the horizontal distance between the two ends of the bridge)

c) How tall is this bridge (Meaning the vertical height of the bridge)

d) If you walk onto the bridge and then walk forward 10 meters on the bridge, how high off the ground will you be?

a) Enter the distance in meters, do not include units

b) Enter the horizontal length in meters, do not include units

c) Enter the vertical height in meters, do not include units

d) Enter the height you'll be at in meters, do not include units

I don't know

Summary - Factored Form of a Quadratic Equation
Given the factored form of a quadratic equation y=a(x−r)(x−s)y=a(x-r)(x-s)y=a(x−r)(x−s):