Identify the vertex and the axis of symmetry for the graph of y=5(x-2)^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=ax2 + 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 x2 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(x-2)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(2-2)2+3=3
d. y=5(-2-2)2+3=83 (not equal to 3)

To identify the vertex and the axis of symmetry for the graph of the equation y = 5(x-2)^2 + 3, we can use the equation in vertex form, which is y = a(x-h)^2 + k.

In this equation, (h, k) represents the vertex of the parabola, and the axis of symmetry is the vertical line passing through the vertex, given by the equation x = h.

Comparing the given equation y = 5(x-2)^2 + 3 with the vertex form equation, we can see that h = 2 and k = 3.

Therefore, the vertex of the parabola is (2, 3).

The axis of symmetry is the vertical line passing through the vertex, so the equation of the axis of symmetry is x = 2.

So, the correct answer is: c) vertex (2,3); x = 2.

Hope that helps! Let me know if you have any more questions.