y = 16x^2 + 36
Lets find the intercepts, making x and y to zero and solving:
0 = 16x^2 + 36
16x^2 = -36
therefore there is not x-intercept, that is, when y equals zero the function does not have a value, is not defined.
Now, the y intercept, that is when x equals zero:
y = 16x^2 + 36
y = 16(0)^2 + 36
y = 36
so, when x = 0, y = 36, that is, the function hits the y axis when x = 0