Original post by honour
e.g. y = x2 - 4x + 5
Complete the square so it is in the form and then the minimum/maximum point will be at
Original post by honour
e.g. y = x2 - 4x + 5
Why not?
Yes - the roots just dictate the x intercepts, but the the shape of the graph will still remain intact thus the max/min points can be identified.
Original post by solC
Complete the square so it is in the form and then the minimum/maximum point will be at
Thanks, I previously thought this was only possible for quadratics with real roots.
Original post by agrajaag
You could also find the derivative and equate it to 0 and solve for x. This will give you the x value of the min/max point. You can find y by calculating f(x)
I think differentiation might be a bit too much for him at this point xD
Perhaps the fact that you're asking this question suggests that you don't really understand what minimum/maximum points of a quadratic curve are. They are the turning points of the curve (in the case of quadratics) and every quadratic curve has a turning point since if then finding the turning points we solve so and so there is always a (real) solution. You could also complete the square to get the same result.
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