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Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

Cumulative Assessment

Exercise 4 Page 232

Find the equation of the axis of symmetry for all the given points.

See solution.

Practice makes perfect

Let's put all the given points on a coordinate plane.

Notice that these are all on a horizontal line. The axis of symmetry of a parabola through any two of these points is vertical.

Moreover, this axis of symmetry is halfway between the two points, so we can find the equation using the averages of the first coordinates of the two points. Let's find this equation when we pair the fixed ( 3,1) with each of the given points.

Points	Average of First Coordinates	Axis of Symmetry
( - 5,1) and ( 3,1)	- 5+ 3/2= - 1	x= -1
( - 4,1) and ( 3,1)	- 4+ 3/2= -1/2	x= -1/2
( - 3,1) and ( 3,1)	- 3+ 3/2= 0	x= 0
( - 2,1) and ( 3,1)	- 2+ 3/2= 1/2	x= 1/2
( 0,1) and ( 3,1)	0+ 3/2= 3/2	x= 3/2
( 1,1) and ( 3,1)	1+ 3/2= 2	x= 2

We can read the answer to the questions from the table.

If the axis of symmetry has equation of the form x=- a (for some a>0), the points (-5,1) and (- 4,1) could lie on the parabola.
If the axis of symmetry has equation of the form x=a (for some a>0), the points (-2,1), (0,1), and (1,1) could lie on the parabola.

Subchapter links

Exercises p.232-233