Question

The gradient function of a curve is given by
and the curve y = f(x) passes through the point (0, 12).
(i) Find the equation of the curve y = f(x).
(ii) Sketch the curve y = f(x), clearly labelling turning points and the y intercept.

Answer 1

The gradient function of a curve is given by....

Was there suppose to be something written after that?

Answer 2

yes sorry, the gradient of a curve is given by f ' (x) =3(x+1)(x-3) and... the rest is correct

Answer 3

Oh I think I'm just getting confused by the word "gradient".
I'm used to seeing that in multivariable calculus, but not in single variable equations. So we're given the DERIVATIVE function, and we need to know what the function f(x) is.

So we need to apply anti-differentiation. Or are you familiar with the idea of integration yet? :)

Answer 4

yes i am

Answer 5

So we'll want to expand these factors, then we find the anti-derivative fairly easily, applying the power rule for integration to each term.

Answer 6

\[\huge f'(x)=3x^2-6x-9\]I think we get something like this when we expand it out.

Answer 7

do we expand the brackets then times all the numbers by 3 ?

Answer 8

Yah :)

Answer 9

so would it be x^3-3x^2-9x

Answer 10

\[\huge f'(x)=3(x+1)(x-3)\]Multiplying out the 2 factors gives us,\[\huge f'(x)=3(x^2+x-3x-3)\]Combining like terms,\[\huge f'(x)=3(x^2-2x-3)\]Multiplying everything by 3 gives us,\[\huge f'(x)=3x^2-6x-9\]
Having trouble on the expansion part?

Answer 11

nope thanks ! :)

Answer 12

Understand the power rule for integration alright? :D
It's like the power rule for differentiation but backwards, and in the reverse order.
Just need to remember that CONSTANT C that appears when we anti-differentiate.

Answer 13

oh so it would be x^3-3x^2-9x+C

Answer 14

Yesss very good :D

Answer 15

Then you use the information they gave you about the point it passes through, plug it into the equation, and solve for C.

Answer 16

so is this still f '(x) or is it f(x) ?

Answer 17

\[\huge f'(x)=3x^2-6x-9\]

\[\huge f(x)=x^3-3x^2-9x+C\]

Answer 18

f(0)+x^3-3x^2-9x+C

Answer 19

f(0)=*

Answer 20

\[\large f(0)=12 \qquad \rightarrow \qquad 12=0^3-3\cdot0^2-9\cdot0+C\]See how that works?

Answer 21

12=c

Answer 22

K looks good :D so rewrite f(x) with your new found C value.

Answer 23

f(x) = x^3-3x^2-9x+12 thanks !!

Answer 24

Yay team \c:/