quick questions help :)

[hello_kitty]

Hey Guys,
I just have a few questions im confused on how to approach

1) The curve with the equation y = ax^2 + bx has a gradient of 3 at the point (2,-2)
a) Find the values of a and b

For this it tried using hte derivative 2ax + b then subing in both 2 and -2..then working it simultaneously but it didnt work.it was just a try :/

The anser is a = 2, b = -5
b) Find the coordinates of the point where the gradient is 0?

Not sure how to go about this, is it when i find the gradient i work out the derivative then equal it to 0??

answer:  (5/4, -25/8)



2a) Let P be the point (1, f(1)) and Q the point (1 + h, f(1+h)). Find the gradient of the chord PQ.

* before this question it asked to sketch the graph f(x) = 1/x^2  (TRUNCAS), x does not equal 0  - I've done this already, just thought it might be informationt hat is needed


Doesnt finding the chord mean find the derivative? but how do i work it out? ive got (f(1 + h) - f(1)) / 1+h-1

The answer is -2 - h/(1+h)^2


b) Hence find the gradient of the curve f(x) = 1/x^2 at x = 1


Thanks!!!

[tony3272]

for 1a)
The simultaneous equations you should have are -2=4a+2b and 3=4a + b
so you rearrange the second to b= 3-4a
which you then sub into the first equation
so -2=4a+2(3-4a)
-2=4a+6-8a
therefore -4a=-8
therefore a=2
and hence b=-5

For 1b) i think you forgot the x which is multiplied by the b.
you derivative equation will be dy/dx= 4x-5
which your x coordinate will then by x=5/4
and you then get y=-25/8 when you sub it into the original equation.

Lol sorry i thought you wrote that you kept getting x=0 when you let the derivative to equal 0. But yeah you just derive the function and let it equal 0 to work out this point

let


at ,



For 2a) your method is correct so just expand it out and simplify.