Vertex of parabola given x intercept and leading coefficient
- Thread starter azws12
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
Re: Vertex of parabola given x intercept and leading coeffic
Well, do it. What's your first plan?
RBGTHGANH
Are we assuming the axis is parallel to either the x-axis or the y-axis?
Two x-intercepts, so the axis must be by parallel to the y-axis.
Leading coefficient < 0, so it opens down.
x-intercepts at (3,0) and (15,0), so vertex must be at x = (3+15)/2 = 9
That's quite a bit of information. Let's see what you get.
Well, do it. What's your first plan?
RBGTHGANH
Are we assuming the axis is parallel to either the x-axis or the y-axis?
Two x-intercepts, so the axis must be by parallel to the y-axis.
Leading coefficient < 0, so it opens down.
x-intercepts at (3,0) and (15,0), so vertex must be at x = (3+15)/2 = 9
That's quite a bit of information. Let's see what you get.