Find the equation of the tangent to the curve defined by that is parallel to the line represented by 3x-y-9=0
This is my work, can someone check it please? I'm not sure if i have the rules of the natural exponential function correct.
First off I want to say, I know that
BUT
is there anything for
Now I will continue.
y'=3 at x=?
x=-lne/ln2
x=-1/ln2
y=mx+b
y=mx+b
3(-1/ln2)+b=
b=
Final answer:
y=3x+
ok.. why do we have to use that here? if I was to use y=mx+b (my teacher uses this in class) how would it look? I am trying to figure it out using that way
my work:
y=mx+b
-(ln2+2)=3(-ln2)+b
-ln2-2=-3ln2+b
-ln2+3ln2-2=b
-ln2+ln8-2=b
ln6-2=b
edit: if this is still wrong, can you please show me using the y=mx+b formula? thanks