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What is the coordinate for the vertex of the parabola represented by the equation y equals x2-6x plus 16?
Wiki User
∙ 14y ago
The parabola y=x2-6x+16 is an upward-opening parabola. Although with most parabolas in most math classes such a parabola would be a perfect square and factor down to the form y=(x-a)2 where "a" is the x-coordinate of the vertex, for this parabola this is not true.
So, to find the vertex, I will employ derivative techniques from basic calculus (since this is the "calculus" section, I assume you know how to use these). Since the function is an upward-opening parabola, its only critical point will be a minimum at its vertex. So, to find the vertex we must find the minimum.
f(x)=x2-6x+16
f'(x)=2x-6
2x-6=0 ---> 2x=6 ---> x=3
Although we know this will be a minimum, it doesn't hurt to confirm that it is using the first derivative test.
0<3
f'(0)=-6
5>3
f'(5)=4
Since the values of f'(x) are negative when x is less than 3, zero when x is 3, and positive when x is greater than 3, x=3 is a minimum.
Therefore, the x-coordinate of the vertex is x=3. Plugging this x into the original equation yields:
f(3)=(3)2-6(3)+16=9-18+16=7
So the vertex of the parabola is (3,7)
Wiki User
∙ 14y ago
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