Graph question (1 Viewer)

[BlueGas]

I need someone to explain the answer for me. I knew that there was a point of inflexion at -2 because it was a stationary point at f'(x), and I also knew that max at x = -1 and a min at x = 0 because the parabola cuts these points so these turning points turn into stationary points, even though I knew it was a cubic, I still got my sketch wrong because I didn't know the max occurs at (-1, 1) and the min occurs at (0, -1), my question is how am I meant to know that max and min curves go on y= 1 and y= -1?

 

[BlueGas]

Bump.

 

[InteGrand]

I need someone to explain the answer for me. I knew that there was a point of inflexion at -2 because it was a stationary point at f'(x), and I also knew that max at x = -1 and a min at x = 0 because the parabola cuts these points so these turning points turn into stationary points, even though I knew it was a cubic, I still got my sketch wrong because I didn't know the max occurs at (-1, 1) and the min occurs at (0, -1), my question is how am I meant to know that max and min curves go on y= 1 and y= -1?

 

[BlueGas]

Yeah I sketched the maximum and minimum. I drew the maximum similar to answer but I drew the minimum at (0,0), and as you can see it's different in the answer. Would I still get my sketch wrong because I drew the minimum at (0,0)?

 

[InteGrand]

Yeah I sketched the maximum and minimum. I drew the maximum similar to answer but I drew the minimum at (0,0), and as you can see it's different in the answer. Would I still get my sketch wrong because I drew the minimum at (0,0)?

I think so, there's nothing that allows us to assume it's the minimum, and also, you should draw the curve symmetric about its point of inflexion, since all cubics are symmetric about their points of inflexion. Basically, the max. and min. should have arbitrarily drawn y-values since we don't know (or care, for this question) what they are, and by drawing it at the origin we are assuming we know what it is, which we can't do.

You could easily derive the equation of the cubic from the given graphs of the derivatives if you wanted, and then find the values of the extrema, but this is unnecessary and a waste of time.

 

[BlueGas]

I think so, there's nothing that allows us to assume it's the minimum, and also, you should draw the curve symmetric about its point of inflexion, since all cubics are symmetric about their points of inflexion. Basically, the max. and min. should have arbitrarily drawn y-values since we don't know (or care, for this question) what they are, and by drawing it at the origin we are assuming we know what it is, which we can't do.

You could easily derive the equation of the cubic from the given graphs of the derivatives if you wanted, and then find the values of the extrema, but this is unnecessary and a waste of time.

So how would I know that the maximum is at (-4,2) and the minimum at (0, -1)?

 

[InteGrand]

So how would I know that the maximum is at (-4,2) and the minimum at (0, -1)?

You don't need to know the y-values of the local extrema to get the sketch for this question. (Notice how the solutions didn't specify them in their sketch.)

 

[BlueGas]

You don't need to know the y-values of the local extrema to get the sketch for this question.

What I mean is that how am I meant to know where to position the maximum and minimum?

 

[InteGrand]

What I mean is that how am I meant to know where to position the maximum and minimum?

You already know the x-values of them and their nature (i.e. which one is max. and which one is min.); this is all you need to know. Just put them arbitrarily somewhere, like the solutions did. Put the local max. above the x-axis and the local min. below it, and the inflexion point is directly between. Note that the shape of the cubic is with y going to +infinity as x goes to +infinity, because its derivative is an upwards facing parabola.

 

[InteGrand]

As long as you get the shape right, you should be fine; since they didn't give you a point that the graph of y = f(x) passes through, the possible graphs of y = f(x) will be a family of curves with the same shape but different vertical shifts, due to the fact that they can differ by a constant (since when you integrate the f' function, you get a +C). So your y-values are unimportant. Just need the right shape and x-values.

 

[BlueGas]

I drew the graph like this when I originally did this question, would I have gotten it right?

 

[InteGrand]

I drew the graph like this when I originally did this question, would I have gotten it right?

I'm not sure; that is certainly a possible graph of y = f(x). But it is probably best not to show it directly through the origin, to leave it more general (unless they told you it goes through the origin – remember, there's infinitely many possible graphs, all differing by a constant, if we're not told a point that the curve passes through). You should draw a dotted line from x = -4 to the local max. like they did in the solutions though, and one from x = -2 to the inflexion point, and label all these key points (basically do what the solution did for their sketch of y = f(x)).

 

[BlueGas]

I'm not sure; that is certainly a possible graph of y = f(x). But it is probably best not to show it directly through the origin, to leave it more general (unless they told you it goes through the origin – remember, there's infinitely many possible graphs, all differing by a constant, if we're not told a point that the curve passes through). You should draw a dotted line from x = -4 to the local max. like they did in the solutions though, and one from x = -2 to the inflexion point, and label all these key points (basically do what the solution did for their sketch of y = f(x)).

So I would still earn marks for drawing this graph right?

 

[InteGrand]

So I would still earn marks for drawing this graph right?

I think so. Check the marking criteria if they are available.

 

[BlueGas]

Didn't want to create a new thread for this but I have a quick question, if there was for example, a parabola with a stationary point at x = 1, that's it, nothing else is given in the question, no equation, just a stationary point (minimum) at x = 1, how would I draw the derivative of this? How would I know if the derivative is a positive or a negative gradient?

 

[InteGrand]

Didn't want to create a new thread for this but I have a quick question, if there was for example, a parabola with a stationary point at x = 1, that's it, nothing else is given in the question, no equation, just a stationary point (minimum) at x = 1, how would I draw the derivative of this? How would I know if the derivative is a positive or a negative gradient?