The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
Answer:
Here is your answer with solutions.
Answer:
.2
Step-by-step explanation:
do 1 divided into 5 in long division.
you will end up with .2
Y=-x+2
OR if you are looking for point slope:
y+1=-1(x-3)
Answer:
3
Step-by-step explanation:
just do (-1/2) x (-6) where the negatives cancels out and half of 6 is 3
Brainilist Maybe?
