finding an equation of the tangent line to a curve at the given point in MATLAB

I tried looking for and solving this for an hour and I am stumped.

the question I am trying to have answered is: find an equation of the tangent line to the curve at the given point y=4x-3x^2, (2,-4)

Answer 1

[Birdman]

Actually, it is quite simple to do. For instance, let's define the equation symbolically in MATLAB:

syms y(x)
y(x)=4*x-3*x^2;

Then, take the derivative:

dy(x)=diff(y,x)

which would be

4-6*x

At this point, you can find the slope of the tangent line at point (2,-4) by inserting 2 into the above equation, which would be

4-6*(2)=-8

You know that the slope of tangent line is -8, but you should also find the value of y for that tangent line. Consider it as

y=-8*x+c

Then find the c by simply

-4=-8*(2)+c

where c would be 12.

If you plot both curves in the same figure, you will get

where the blue line is the original equation and red line is tangent line.

Answer 2

[KAMEWSARA]

y=x.^2