[StephenP15]

Hi, I have to teach myself C1 for edexcel this holiday and I'm really stuck with sketching curves, because I don't know which shape the curve should go. One of the questions was y=x^3+5x^2+4x and I was able to factorise it to get the co-ordinates (0,0) (-4,0) (-1,0), but I'm not sure which way to draw the curve, whether it has a starting positive or negative gradient? Sorry to ramble the textbook isn't very helpful!

[jameswhughes]

(Original post by StephenP15)
Hi, I have to teach myself C1 for edexcel this holiday and I'm really stuck with sketching curves, because I don't know which shape the curve should go. One of the questions was y=x^3+5x^2+4x and I was able to factorise it to get the co-ordinates (0,0) (-4,0) (-1,0), but I'm not sure which way to draw the curve, whether it has a starting positive or negative gradient? Sorry to ramble the textbook isn't very helpful!

Look at the values for positive and negative infinity, e.g as the leading term is you can see that for large positive values of x the value of y will also be large and positive.

[tehforum]

Just remember that for an x^3 graph, it ALWAYS start from bottom left, to top right.

[crazycake93]

(Original post by StephenP15)
Hi, I have to teach myself C1 for edexcel this holiday and I'm really stuck with sketching curves, because I don't know which shape the curve should go. One of the questions was y=x^3+5x^2+4x and I was able to factorise it to get the co-ordinates (0,0) (-4,0) (-1,0), but I'm not sure which way to draw the curve, whether it has a starting positive or negative gradient? Sorry to ramble the textbook isn't very helpful!

If you don't know which way to sketch the curve, substitute random numbers in and actually plot the graph. It always helps in an exam.

[Lukedavidhopkins1]

(Original post by tehforum)
Just remember that for an x^3 graph, it ALWAYS start from bottom left, to top right.

Not if the negatives are in particular places so as to make it a reflection in the y or x axis

[StephenP15]

Thank-you all so very much I get it now The Heinemann textbooks are really annoying because they just mention x and infinity haha

[StephenP15]

Another little quick question, I'm looking at transformations which has the quesstion. Sketch the curves: y= f(X) where f(x)= (X-1)(x+2) and then on seperate diagrams sketch i) y=f(X+2) ii y=f(X)+2. I was able to do the sketches, but the next question was really confusing, find the equations of the curves y= f(x+2) and y=f(X)+2, in terms of x, and use these equations to find the coordinates of the points where your graphs in part b cross the y-axis. Sorry it's a long question I'm just a bit stuck, as the example question hasn't helped!

[baphomet]

To 'create' the equation of f(x+2) from f(x)

replace each x with (x+2) in the f(x) you were given; it is best to keep the brackets as good practice

in other words f(x) = (x-1)(x+2) was given so f(x+2) = ( (x+2) - 1) ( (x+2) + 2) as stated each x has been substituted with a (x+2)

You can then simplify this. The graph of this new function crosses the y-axis when x=0 or simply look at the constant produced which is the y-intercept value exactly like y=mx+c work.

For f(x) + 2 you create this by simply adding on 2 to the original f(x) and simplifying again. You are best simplifying to ax^2 + bx +c form.

[StephenP15]

Thank you very much!! It's just hard to click when you have no one to explain it!