Everything in Quadratics!! | In Just 40 Minutes | Chapter 2 Pure Maths | Year 1/AS Level Revision

[@leejohnson197733]

this is great but im still a bit lost, im trying to look at y = 7x^2 - 3x how do i find the co ordinates and maxlimu / min

[@ImStuckRevision]

You will need to use differentiation to do so. So differentiate y, then set this function = 0 then solve for x, this will give you the x value for the minimum/maximum

[@leejohnson197733]

@@ImStuckRevision ok so would that be 2(7x^1 - 3x)

[@ImStuckRevision]

Not quite, 14x - 3 as you must treat both terms separately, so by multiplying by the power and reducing it by one would make 7x^2 into 14x And -3x or -3x^1 would differentiate into -3

[@leejohnson197733]

@@ImStuckRevision ok i think i see what you mean, so does this answer give you the turning point? then i have to go on to find max and minima

[@ImStuckRevision]

Yes the turning points are also minimum and maximum. To find out if they are either minimum or maximum you must differentiate again.