Are these equal?

.... i don't really get what to do???

i'm not getting the bit where i do something with the gradient and find the co-ordinates of the point of intersection.

It's not clear which part you're stuck at. Can you point to the exact part of my last post where you get stuck / something is unclear?

Are you able to solve the question that I wrote in italics?

Initially I gave you this other method for you to look at and ignore if you like. But I think it's important for your understanding that you are able to follow this method.

i got k as -4???? but that can't be right

yes from the bit in italics i did this

$....[br][br]11=\dfrac{1}{3} x^2 +8 \\ 33=x^2 +8[br]...[br]$

i got k as -4???? but that can't be right

yes from the bit in italics i did this

$y=\dfrac{1}{3} \times 9 +8[br][br]y=11[br][br]11=\dfrac{1}{3} x^2 +8[br][br]33=x^2 +8[br][br]x^2=25[br][br]x=5[br][br]11=15+k[br][br]k=-4$

11=\frac{1}{3} x^2 +8 \nRightarrow 33=x^2 +8

You've substituted $x=3$ into the curve equation. Why? You should be thinking about why you're doing each step.


The gradient of the curve at P is 3. Do you understand this part? Please tell us and we can explain more if you don't.

If the gradient of the curve at a point is 3 then this means that $\frac{dy}{dx}=3$ at the point.

$\displaystyle y=\frac{1}{3}x^2+8 \Rightarrow \frac{dy}{dx} = \frac{2}{3}x$

Again, tell us if you don't understand this.

$\displaystyle \frac{dy}{dx} = 3 \Rightarrow \frac{2}{3}x = 3$

If you solve this you will have the x-coordinate of P.

it happened earlier above too >.> i need to be more careful......



I do understand what that means, at point P on the curve C the gradient is 3

why do i equate gradients??? what does that do?

Can you tell me where I have equated gradients?

It's easier for me to understand if you say the specific part of my post that confuses you.

where you put the gradient of

$y=3x+k[br]and[br][br]y=\dfrac{1}{3} x^2 +8$

together....

where you put the gradient of

$y=3x+k[br]and[br][br]y=\dfrac{1}{3} x^2 +8$

together....

The gradient of the parabola is constantly changing but as the straight line is a tangent you know that it hits the curve at one point, P. At that point, the gradient of the parabola is equal to the gradient of the straight line. This allows us to work out the coordinate of the point of intersection and by extension, k.

Please quote the part of my post where this happened.

I'm finding it hard to pinpoint what you're unsure about.

I'd appreciate if someone else has a go at explaining this - it's often good to hear two explanations

Please quote the part of my post where this happened.