A certain quadratic function has x-intercepts at 2 and 3. What are the x-coordinates of its vertex?
2 answers:
Answer:
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
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