Question 551723: y = (1/16)x^2
how do i solve this to find the vertex, focus, equation of diretrex, axis of symmerty, and the x and y intercepts ?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y = (1/16)x^2
how do i solve this to find the vertex, focus, equation of diretrex, axis of symmerty, and the x and y intercepts ?
**
y = (1/16)x^2
x^2=16y
This is an equation for a parabola of standard form: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex.
For given equation:
vertex: (0,0)
axis of symmetry: x=0 or y-axis (parabola opens upwards)
4p=16
p=4
focus: (0,4) (4 units above vertex on the axis of symmetry)
directrix: y=-4 (4 units below vertex on the axis of symmetry)
x-intercept=0
y-intercept=0
|
|
|
