Choose the correct x-intercepts for the following quadratic function. a. x = 1, x = 3 b. x = 4, x = –3 c. x = 2, x = –7 d. x = 2, x = –1

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We can still have some fun with answer choice  x = 1, x = 3:

An associated quadratic function would have the form y = a(x-1)(x-3).  Just supposing that the point (2, 5) were on the graph, then this   y = a(x-1)(x-3) would become   5 = a(2-1)(2-3), or 5 = a(1)(-1), or 5 = -a, or a = -5.

Thus, the equation of the parabola would be y = -5(x-1)(x-3), or, in the more usual form, y = -5(x^2 - 4x + 3).

Complete the square to find the vertex:

Steal y = x^2 - 4x + 3 for a moment and complete the square:

x^2 - 4x + 3    =  x^2 - 4x + 4 - 4 + 3, or (x-2)^2  -  1

Subbing this back into    y = -5(x^2 - 4x + 3), we get

y = -5 [ (x-2)^2  -1 ], or   y = -5(x-2)^2 + 5.  This shows that the vertex is at (2, 5), that the graph opens downward, and the graph is symm. about the line x = 2.  

Having fun yet?