Sorry, I was making the point that there's no reason to wait until Year 12 to teach it because of how basic it is. But whatever, doesn't really matter either way.Aplus said:
LOL, how come you're not teaching him?jellybelly59 said:
What the hell was I supposed to do if my school doesn't teach it in year 11? So I just learnt it myself.Continuum said:Sorry, I was making the point that there's no reason to wait until Year 12 to teach it because of how basic it is. But whatever, doesn't really matter either way.
LOL, how come you're not teaching him?
Yeah, that's the method I used for most of the questions from the Year 11 3 Unit Cambridge Chapter 8, because I hadn't learnt Differentiation at that time.3unitz said:you can find the gradient of the tangent without differentiation, its just longer:
let y = mx + c be the equation of the tangent which touches parabola at point (-2, 5)
equate to find general equation for point of intersection of line and parabola:
mx + c = x^2-2x-3
x^2 - x(2 + m) - (3 + c) = 0
since we're after a tangent, the line only touches the parabola once, therefore discriminant = 0:
(2 + m)^2 + 4(3 + c) = 0 ---------(1)
since (-2,5) lies on the line y = mx + c we know:
5 = -2m + c
c = 5 + 2m ------(2)
sub (2) into (1):
(2 + m)^2 + 4(3 + 5 + 2m) = 0
4 + 4m + m^2 + 32 + 8m = 0
m^2 + 12m + 36 = 0
(m + 6)^2 = 0
m = -6
due to the symmetry the gradient of the other tangent is +6
how fucking awesome is that!!!
Yeh me too. And you know what's stupid, schools are called good if they offer accelerated. And you know what's stupider, people judge how smart themselves are by what school they go to. Arrogant fools!rolror88 said:I have no gun