By completing the square, find any turning points and intersects with the x and y axes of the following curve. f(x) = 2x^2 - 12x +7
f(x) = 2x2 - 12x +7
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First take out a factor of 2 so that we have the coefficient of x2 as 1. f(x) = 2[x2 - 6x +7/2]
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Next, complete the sqaure on the part in square brackets. f(x) = 2[(x - 3)2 - 9 + 7/2]
f(x) = 2[(x - 3)2 - 11/2]
f(x) = 2(x - 3)2 -11
- We know from the fact that the original equation had a positive coefficient of x2 that the turning point will be a minimum. Therefore, in order to find the turning point we must minimise the equation we have arrived at. It can be seen that the lowest value of f(x) possible will occur when x = 3, giving the turning point as:
(3, -11)
- In order to find the intersection points with the x axis, we must set y = 0.
f(x) = 2(x - 3)2 - 11 = 0
2(x - 3)2 = 11
(x - 3)2 = 11/2
(x - 3) = +/- sqrt(11/2)
x = 3 +sqrt(11/2) or x = 3 - sqrt(11/2)
-This gives the intersection points as:
(3 + sqrt(11/2), 0) and
(3 - sqrt(11/2), 0)
- In order to find the intersection point with the y axis, we must set x = 0.
f(x) = 2(0 - 3)2 - 11
f(x) = 2(9) -11
f(x) = 7
- This gives the intersection point as:
(0, 7)
