Partial Factoring (Lesson).notebook
WARM-UP
a) y = ­x2 + 6x + 1
b) y = 0.5x2 ­ 0.6x
HOMEWORK TAKE­UP
Pg. 116 #2,3
UNIT #3: Quadratic Functions
Partial Factoring
Lesson: Partial Factoring
Learning Goal:
I will learn how to use partial factoring to find the vertex of the
parabola.
Partial Factoring involves finding two points
on the parabola that have the same y‐coordinate.
We know that if we have two points on a parabola
that have the same y‐coordinate, we can find the
x‐coordinate of the vertex by taking the average
of the two coordinates.
Partial Factoring (Lesson).notebook
Steps to Follow When Partial Factoring
Step 1: Look at constant term in the expression. Take
away that amount from each side of theexpression.
Step 2: Factor whatever is left in the expression, and
find the two places where x is equal to zero.
NOTE: These are NOT the roots. These x‐coordinates
have the same y‐ coordinate.
Step 3: Find the Axis of Symmetry of the parabola. We
can do this by taking the average of thetwo x‐
coordinates where the y‐values are the same. Now, we
know the x‐value of the vertex.
Step 4: Substitute the x‐value of the vertex, into the
original equation to find the y‐value of the vertex.
Example continued:
Now, find the average of the two x‐coordinates
to find the Axis of Symmetryof the parabola.
(0 + (‐2)) ÷ 2 = ‐1
Step 3: Find the axis of
symmetry by finding the
average of two x coordinates.
Example:
Use Partial Factoring to find the vertex ofthe
parabola with the equation y = x² + 2x ‐ 35
y = x² + 2x ‐ 35
Let y = ‐35
‐35 = x² + 2x ‐ 35
‐35 + 35 = x2 + 2
0 = x2 + 2x
0 = x² + 2x
0 = x (x + 2)
x = 0 or x = ‐2.
Step 1: Take away constant
term from each side of the
expression
Step 2: Factor whatever is
left and find where x = 0
When the parabola has a y‐coordinate of ‐35,
the correspondingx‐coordinates are 0 and ‐2.
Factoring to Determine the Zeros
We use completing the squareor partial factoring to
determine the vertex of the parabola.
Therefore, the vertex is at the point (‐1,
y)
We now have to find out what thematchingy coordinate
is for the vertex.
When x = ‐1,
y = (‐1)² + 2(‐1) ‐ 35
y = 1 ‐ 2 ‐ 35
y = ‐36
Step 4: Substitute x­value of
vertex into original equation.
Therefore, the vertex is at the point (‐1,‐36)
We factor the quadratic equationto determine where
the parabola crosses the x ‐ axis.
Partial Factoring (Lesson).notebook
Example continued....
Factor
to determine where the parabola crosses the
x ‐ axis.
y = x 2 +2 x ‐ 35
Plot the zeros, vertex
and points obtained
from partial factoring.
zeros (x‐intercepts)
(‐7, 0) and (5,0)
(‐2, ‐35)
The parabola crosses thex ‐ axis at x = ‐7, and x = 5
(0, ‐35)
vertex:
(‐1, ‐36)
UNIT 3: Quadratic Functions
Partial Factoring
Learning Goal:
I will learn how to use partial factoring to find the vertex of the
parabola.
y = x 2 +2 x ‐ 35
Success Criteria:
Draw a smooth curve
through those
5 points.
To be successful, I must be able to....
• Find the corresponding x values of y, when y is a set as the
constant value for a quadratic equation
• Find the vertex of the parabola using partial factoring
• Find the zeros of the parabola using factoring
• Sketch the parabola using the five points
Practice Work
Axis of
Symmetry
Pg. 116 #2 (every other)
Find the vertex of the parabolas using
partial factoring.
#2f ‐ Find the vertex by partial factoring and the
zeros by factoring
Sketch the parabola for 2f.