Let us find the vertex point (h, k)
h is the x-coordinate of the vertex point of the equation
[tex]y=ax^2+bx+c[/tex][tex]h=\frac{-b}{2a}[/tex]
a is the coefficient of x^2,
b is the coefficient of x
From the given parabola
a = 1/4
b = 0
So
[tex]h=\frac{0}{\frac{1}{4}}=0[/tex]
h = 0
k is the y-coordinate of the vertex point, to find it substitute x by h in the equation
[tex]y=\frac{1}{4}x^2[/tex]
Substitute x by h = 0
[tex]\begin{gathered} k=\frac{1}{4}(0)^2 \\ k=0 \end{gathered}[/tex]
Then the vertex point is (0, 0) ==== the origin point
To find two points on the graph right the vertex point substitute x by 2 positive values greater than 0
We can choose 4 and 8
at x = 4 ===== y = 1/4 (4)^2 = 1/4(16) = 4
The first point on the right side is (4 , 4)
at x = 8 ===== y = 1/4(8)^2 = 1/4(64) = 16
The second point is (8, 16)
We will plot the other points on the left with x = -4 and -8
It will give us the same values of y with the first two points
(-4 , 4) and (-8, 16)
That because (-4)^2 = 16 like (4)^2 = 16
The even powers cancel the negative signs
So we have 5 points to draw the graph
(-8 , 16) , (-4 , 4) , (0, 0) , (4 ,4), (8 , 16)
