Question

State the vertex and axis of symmetry of the graph of y=ax2+c.

Answer 1

in vertex form

$y=a(x-h)^2+k$

the vertex is (h,k) and the axis of symmetry is x=h

put into vertex form

$y=ax^2+c$

$y=a(x)^2+c$

$y=a(x-0)^2+c$

h=0, k=c

the vertex is (0,c)

the axis of symmetry is x=0

Answer 2

State the vertex and axis of symmetry of the graph of y=ax^2+c

General form of quadratic equation is $y=ax^2 + bx +c$

There is no bx in our given equation, so we put 0x

Given equation can be written as $y=ax^2 + 0x +c$

a=a , b=0

Now we use formula to find vertex

$x=\frac{-b}{2a}$

$x=\frac{-0}{2a}=0$

Now we plug in 0 for 'a' and find out y

$y=a(0)^2 + 0x +c= c$

So our vertex is (0,c)

The axis of symmetry at x coordinate of vertex

So x=0 is our axis of symmetry