Question 468040
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: 
   {{{x = a(y-k)^2 + h}}} where (h,k) is vertex
- Use formula {{{y=(-b)/2a}}} where {{{x= ay^2 + by + c}}} 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.
{{{x = 4y^2 +9y +10}}}
a = 4
b = 9
c = 10
{{{y=(-b)/2a = (-9)/8}}}
Substitute this back in and solve for x
{{{x = 4(-9/8)^2 +9(-9/8) +10}}}
{{{x = 4(81/64) -(81/8) +10}}}
{{{x = (81/16) -(81/8) +10}}}
{{{x = (81/16) -(162/16) +(160/16)}}}
{{{x = (81-162+160)/16 = 79/16}}}
Vertex: ({{{79/16}}},{{{(-9)/8}}})
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 {{{9/8}}} in other direction x will also be 10
(10,{{{(-18)/8}}})
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**
{{{graph(300,300,-6,4,0,12,4x^2+9x+10)}}}