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:
<u>m = 0</u>, y = -6
Step-by-step explanation:
using point slope form :
y-y1=m(x-x1) ; initial point (x1,y1) = (4,-6), final point (x,y) = (7,-6).
one substituted, it should look like this:
-6--6 = m(7-4),
-6+6= m(7-4)
0 = m(7-4)
0 = 3m
0/3 = 3/3m
0 = m
m = 0/3 = 0
Answer:
f(1) = √(4*1) = √4 = 2
The point (1, 2) is only on the third graph.
Answer:
Sarah mowed 1 lawn in 3 hours
Step-by-step explanation:
3 lawns divided by 3 is 1 lawn
9 hours divided by 3 is 3 hours
