First, notice that the variable raised to the power of 2 is x. Therefore, the axis of symmetry will be a vertical line. We will start by writing the given equation in the form y=a(x-h)^2+k. To do so, we will first rewrite the equation into having 1 as coefficient for the quadratic term.
Let's now simplify the right-hand side by completing the square. We have to add and subtract ( b2)^2. In this case, we have that the linearcoefficient b is 8.
b=8 ⇒ (b/2)^2=(8/2)^2=4^2
Now, let's identify the values of a, h, and k.
y=3(x+4)^2-58
⇕
y= 3(x-( - 4))^2+( - 58)
We can see that a= 3, h= - 4, and that k= - 58. Next, we can use this information to highlight some important characteristics of the parabola.
y=a(x-h)^2+k
y=3(x-(- 4))^2+(-58)
Direction of Opening
Up if a>0, Down if a<0
Up ( 3>0)
Vertex
( h, k)
( - 4, - 58)
Axis of Symmetry
x= h
x= - 4
Focus
( h, k+1/4 a)
( - 4, - 58+1/4( 3)) ⇕ (- 4,- 57 1112)
Directrix
y= k-1/4 a
y= - 58-1/4( 3) ⇕ y=- 58 112
Length of Latus Rectum
|1/a| units
|1/3| units ⇕ 1/3 units
Since the vertex and focus are very close to each other, and the length of latus rectum is very short, it is hard to draw the parabola using this information. Therefore, we need to find a few more points from this graph.
x
y=3(x+4)^2-58
y
- 8
y=3( - 8+4)^2-58
- 10
- 6
y=3( - 6+4)^2-58
- 46
- 2
y=3( - 2+4)^2-58
- 46
0
y=3( 0+4)^2-58
- 10
We can use these points together with the vertex to draw the graph.
