Hi,
the vertex form of a parabola opening right or left, where(h,k) is the vertex.
4y^2 +9y+10 = x
4[(y + 9/8)^2 - 81/64] + 10
4(y + 9/8)^2 - 81/16 + 10 = x
4(y + 9/8)^2 + 79/16 = x V(79/16,-9/8) Opening to the right
Answer by ccs2011(207) (Show Source): You can put this solution on YOUR website! This is a horizontal or sideways parabola because it is solved for x and the y is squared.
To graph this first determine vertex then a 2nd point.
To determine the vertex of this parabola there are 2 methods:
- Use completing the square to put it in vertex form: where (h,k) is vertex
- Use formula where to determine line of symmetry then substitute this value into equation to determine x_value of vertex.
I will use the 2nd since it is the more straightforward and I don't know how familiar you are with completing the square.
a = 4
b = 9
c = 10
Substitute this back in and solve for x
Vertex: (,)
Now we need one other point to graph this parabola
Lets look at x_intercept, if y=0 then x =10, (10,0)
A parabola is symmetric so if we go in other direction x will also be 10
(10,)
Now plot these 3 points and draw a parabola that goes through all 3 points
Here is a graph:
**Imagine its rotated clockwise 90 degrees, it wont do horizontal parabolas**