State the vertex and axis of symmetry of the graph of y=ax^2+c
General form of quadratic equation is [tex]y=ax^2 + bx +c[/tex]
There is no bx in our given equation, so we put 0x
Given equation can be written as [tex]y=ax^2 + 0x +c[/tex]
a=a , b=0
Now we use formula to find vertex
[tex]x=\frac{-b}{2a}[/tex]
[tex]x=\frac{-0}{2a}=0[/tex]
Now we plug in 0 for 'a' and find out y
[tex]y=a(0)^2 + 0x +c= c[/tex]
So our vertex is (0,c)
The axis of symmetry at x coordinate of vertex
So x=0 is our axis of symmetry
The vertex of the graph is (0,c), and the axis of symmetry is [tex]x = 0[/tex]
The function of the graph is given as:
The x-coordinate of the vertex is calculated as:
In [tex]y=ax^2+c.[/tex], the value of b is 0.
i.e. [tex]b = 0[/tex]
So, we have:
[tex]x = -\frac{0}{2a}[/tex]
[tex]x = 0[/tex]
Substitute 0 for x in [tex]y=ax^2+c.[/tex]
[tex]y = a(0)^ + c[/tex]
[tex]y =c[/tex]
So, the vertex of the graph is (0,c).
Also, we have:
[tex]x = 0[/tex]
The above represents the axis of symmetry
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