Partial Factoring (Lesson).notebook
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 the expression. 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 the two 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 Symmetry of 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 of the 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 corresponding x‐coordinates are 0 and ‐2.
Factoring to Determine the Zeros We use completing the square or 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 the matching y 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 equation to 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 the x ‐ 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.
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 the expression. 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 the two 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 Symmetry of 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 of the 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 corresponding x‐coordinates are 0 and ‐2.
Factoring to Determine the Zeros We use completing the square or 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 the matching y 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 equation to 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 the x ‐ 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.
