differentiation c1

find the y coordinate and the value of the gradient at the point P with x coordinate 1 on the curve with equation y=3+2x-x^2

I did
dy/dx=2ax+b

so -2x+2, I don't know what to do next
do I have to make -2x+2=1
and then solve for x?

the answer is 4,0

You've found the gradient function (dy/dx) for the curve.

You could envisage it like;

Gradient = -2x + 2

You're asked to find the gradient when x=1, can you continue from there?

For the y co-ordinate, i'm sure you can probably do that one also.

so -2(1)+2
therefore gradient is 0?
and to find the y coordinate do I have substitute 0 into the original equation?

find the y coordinate and the value of the gradient at the point P with x coordinate 1 on the curve with equation y=3+2x-x^2

I did
dy/dx=2ax+b

so -2x+2, I don't know what to do next
do I have to make -2x+2=1
and then solve for x?

the answer is 4,0

What does $\frac{dy}{dx}$ tell us? Well, it gives us the gradient at a particular point on a curve. So, first you need to find what $\frac{dy}{dx}$ is equal to, which you have found correctly: $\frac{dy}{dx} (3+2x-x^2) = 2-2x$

Now, you want to find the gradient at the point P, hence you need to substitute the x-coordinate into the differential whenever you see x, so you get:

At point P, $\frac{dy}{dx} = 2-2(1)$, which equals what? (I'm sure you can do that!) This is your gradient at the Point P.

Now, the y-coordinate...well that's easy! Don't make the mistake of thinking $\frac{dy}{dx}$ will give you the y-coordinate, as it doesn't - as you know from above, it gives you the gradient. To find the corresponding y-coordinate when $x=1$ you need to substitute the value of x into the equation of the curve, which I'm sure you can do too.

I hope that helps!



Posted from TSR Mobile

find the gradient of the curve where f(x)=x^3-3x+2 at the point A (-1,4)


I m stuck on this question
wt do I hv to do??

Do you understand that the differentiating to find f'(x) gives you the gradient function?

how to find the points on the curve with y=f(x), where the gradient is zero:
a) f(x)=x^3-9x^2+24x-20

here is wt I did, but its wrong
3x^2-18x=0
then factorise it -> 3x(x-6)
x=0 or x=6
y=0 or y=16

b) f(x)=x^3/2-6x+1

But she doesn't say that her expression is $f'(x)$, instead she says it's $f(x)$, hence wouldn't you have to simply insert the x-coordinate?


Posted from TSR Mobile

It's f(x) for both of them we are trying to find the points when the gradient = 0 when the gradient is =0 $f'(x)=0$

So differentiate and set equal to 0

how to find the points on the curve with y=f(x), where the gradient is zero:
a) f(x)=x^3-9x^2+24x-20

here is wt I did, but its wrong
3x^2-18x=0
then factorise it -> 3x(x-6)
x=0 or x=6
y=0 or y=16

b) f(x)=x^3/2-6x+1

A) f (x)=x^3-9x^2+24x-20
first differentiate: dy/dx=3x^2-18x+24
Since grad equal zero=>3x^2-18x+24=0
factorise and (3x+12)(x-2)
x=-12 x=2
y=put -12 in first equation or y=0



Posted from TSR Mobile

A) f (x)=x^3-9x^2+24x-20
first differentiate: dy/dx=3x^2-18x+24
Since grad equal zero=>3x^2-18x+24=0
factorise and (3x+12)(x-2)
x=-12 x=2
y=put -12 in first equation or y=0



Posted from TSR Mobile

Since $f(x) = x^3-9x^2+24x-20 \rightarrow f'(x) = 3x^2-18x+24$, which you've said correctly. Then you've made it equal to zero, which too is correct.

$3x^2 - 18x+24=0 (when factorised) \rightarrow (3x-12)(x-2)=0$ This is your first mistake: you need a $-12 and -2$ to make the constant. Then you've solved it incorrectly too:

You cannot just change the sign of the numbers; it doesn't work with every factorised polynomial - well, most at A Level. You need to use the proper method and make both of the factors equal to zero:

$3x-12=0 or x-2=0$

$When: 3x-12=0,[br]3x=12,[br]x=4$

$Or when: x-2=0[br]x=2$

So the solutions (stationary points) are: x=2 or x=4. To find their corresponding y-values plug these into $f(x)$.


Posted from TSR Mobile

thnx a lot