A certain quadratic function has x-intercepts at 2 and 3. What are the x-coordinates of its vertex? a)7/2
b)1/2
c)3/2
d)5/2

D

Step-by-step explanation:

The equation is going to by

y = (x - 2)(x - 3)         Multiply to remove the brackets.

y = x^2 - 2x - 3x + 6

y = x^2 - 5x + 6       Now what you are going to do is complete the square.

y = (x^2 - 5x  +  __ ) +6

Take - 5

Divide it by 2 = -5 / 2

and square it = (-5 / 2)^2 = 25/4

Put this result inside the brackets.

y = (x^2 - 5x + 25/4 ) + 6               Now subtract 25/4 after the 6

y = (x^2 - 5x + 25/4) + 6 - 25/4

y = (x - 5/2)^2 - 1/4

The x coordinate of the vertex  is 5/2

You could also do this by using x = -b/2a = - -5/2*1 = 5/2

at least two

step-by-step explanation: