How do you graph #y=x^2- 6x + 8# by plotting points?
1 Answer
Apr 18, 2018
"a" is positive = parabola pointing up
"a" is negative = parabola pointing down
First point:
Vertex x-coordinate = -b/2a
plug that answer back into the equation for "x" and then find "y"
(x,y) is the first coordinate
- set "y" to zero = get x-intercept (use factoring or quadratic equation)
- set "x" to zero = get y-intercept(s)
- make a t-chart with "x" on one side and "y" on the other.
Think of any "x" coordinate and then plug it into the equation to solve for the "y" coordinate.
