State the vertex and axis of symmetry of the graph of y=ax^2+c
General form of quadratic equation is
There is no bx in our given equation, so we put 0x
Given equation can be written as
a=a , b=0
Now we use formula to find vertex
Now we plug in 0 for 'a' and find out y
So our vertex is (0,c)
The axis of symmetry at x coordinate of vertex
So x=0 is our axis of symmetry
Slope-intercept form: y = mx + b [m is the slope, b is the y-intercept, or the y value when x = 0 ---> (0, y)]
To find the equation's y-intercepts, you can rearrange the variables, and isolate/get y by itself
#1
cx + ay = b Subtract cx on both sides
ay = b - cx Divide a on both sides
(0,
) is your answer
Do the same for the rest, and you should get:
#2 (0, )
#3 (0, )
#4 (0, )
