# Thread: Sketching a curve from a function

1. ## Sketching a curve from a function

Hi all,

I have a math exam on Tuesday and I am pretty much set when it comes to the exam papers, with one of the exceptions being the style of question shown below:

A function is defined as y=(x+3)(x+5)(x+8). Sketch this curve on graph paper showing the 3 points where it crosses the X axis (between the x values of -1 and -9).
Not really sure how to approach this one. I tried getting the points by going in to my calculator and going to table mode and putting the function in to f(x) and then starting at -1 and ending at -9 in steps of 1. This just kept giving 'MATH ERROR'.

The next part of the questions asks to :

Find the derivative of y=(x+3)(x+5)(x+8) and plot on the same graph paper as part (a).
I am unsure how to go about it after getting the derivative of the function, which I calculated to be 3x^2+32x+79. I am unsure how to get the points from this derivative.

Could someone be so kind to walk me through this question from start to end. Thanks!

2. ## Re: Sketching a curve from a function

try starting at $-9$ and ending at $-1$

3. ## Re: Sketching a curve from a function

Not really sure how to approach this one. I tried getting the points by going in to my calculator and going to table mode and putting the function in to f(x) and then starting at -1 and ending at -9 in steps of 1. This just kept giving 'MATH ERROR'.
works fine in my calculator

4. ## Re: Sketching a curve from a function Originally Posted by Idea try starting at $-9$ and ending at $-1$
That worked. Thanks.

Could someone please demonstrate how to plot the derivative of the used function on to the graph (derivative of function is 3x^2+32x+79).

5. ## Re: Sketching a curve from a function

$f'(x) = 3x^2+32x+79$

parabola that opens upward

zeros at $x = \dfrac{-32 \pm \sqrt{32^2-4(3)(79)}}{6} = \dfrac{-16 \pm \sqrt{19}}{3}$

parabola vertex at $x = -\dfrac{16}{3}$ and $y = f'\left(-\dfrac{16}{3}\right)$

7. ## Re: Sketching a curve from a function

Thanks, guys.