Algebra
posted by Cassie .
Identify the vertex and the axis of symmetry for the graph of y=5(x2)^2 + 3.
a) vertex (2,3); x = 2
b) vertex (2,3); x = 2
c) vertex (2,3); x = 2
d) vertex (2,3); x = 2
I have no idea how to solve this problem! Please help. Thank you! :)

All parabolas of the form
y=ax^{2} + bx + c
have their axis of symmetry passing through the vertex.
Thus a and b can be eliminated on that basis.
Note that when the x^{2} term is positive, the parabola has a minimum at the vertex.
By inspection, when x=2, the value of y is at its minimum because the 5(x2)^{2} term is zero. Any other value of x will render the term > 0. Thus the vertex is at x=2, where y=3.
Thus the answer is (C).
Check:
c. y=5(22)^{2}+3=3
d. y=5(22)^{2}+3=83 (not equal to 3)