Well, I had to go to bed....

But.... a couple of confusions in your notebook. First, note that when your professor wrote

the quantities in parentheses after the f and the f' (i.e., the xo) is not multiplication. This is the traditional notation for a function argument. Thus you should not have the 7 in the following

Also note that the function name is L[x], not l[x] in your plotting function (I am not sure how you got a plot but perhaps you had defined l[x] somewhere else)--Mathematica parameter names are case-sensitive.

So if you make these changes you get the following plot:

And, my advice rom earlier still stands! Start your homework well before the deadline--perhaps the day it is assigned so that you can bring some of your (valid and smart) questions to your professor, TAs, or other classmates to give your learning process a bit more slack. (And, yes, I know you are busy with lots of classes and homework--it;s not easy!)

Use the equation that your professor supplied. Then, as you did, compute the derivative of f and substitute the value of x0 in that expression. Also put that same value into the expression for f itself and then use these in the linearization equation for L[x]. You do not need to use Solve. Think of the reason for (i.e., the idea behind) the expression for L[x]. And (fatherly advice) try to get your homework started well before the midnight deadline ;-) !

In your latest case you will notice that the point at the top does correspond to a horizontal tangent ((just taking a look at it you see that the slope of the curve there is zero), but your other points are not vertical tangents since the slopes of the curve there is not vertical, it has a slope in both cases. The problem is that you substituted into the x values and solved for y rather than substituted for the y values and solved for x in each case,

Here are those two cases, corrected:

NSolve[(84 x + 6 x^2 + 28 x^3 + 3 x^4 - 288 y - 12 y^2 + 4 y^3 ==  5) /. y -> -4, x]

and

NSolve[(84 x + 6 x^2 + 28 x^3 + 3 x^4 - 288 y - 12 y^2 + 4 y^3 ==  5) /. y -> 6, x]

These give the following points (confirm for yourself):

{{-9.14477, -4}, {-2.97489, -4}}

and

{-9.86928, 6}

so your ListPlot is

ListPlot[{{-9.144774626644757`, -4}, {-2.974889393649998`, -4}, \
{-9.869278763054162`, 6}}, PlotStyle -> {Black, PointSize[Large]}]

Ok, this is what I got. Would you say that is right or do I need to switch the x and y values in the coordinates?

Thank you so much for getting back to me! It was extremely helpful!!!

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