Find an equation of the tangent line to the curve at the given point. y=secx (pi/3,2)
1 answer:
To get the equation of tangent, we need a point on it and its slope.
we already have (x1,y1) = (pi/3,2) on the tangent line.
To get the slope of tangent,
m = derivative of y at the given point.
y = sec x
y' = sec x tan x
at x = pi/3,
y' = sec pi/3 tan pi/3
y'= 2
so the slope of tangent is 2
equation will be
y -y1 = m (x-x1)
y - 2 = 2 (x- pi/3)
Thats your equation of tangent.
you can simplify if you want.
we already have (x1,y1) = (pi/3,2) on the tangent line.
To get the slope of tangent,
m = derivative of y at the given point.
y = sec x
y' = sec x tan x
at x = pi/3,
y' = sec pi/3 tan pi/3
y'= 2
so the slope of tangent is 2
equation will be
y -y1 = m (x-x1)
y - 2 = 2
Thats your equation of tangent.
you can simplify if you want.
You might be interested in
Answer:
g(x) is shifted 6 units to the left
Step-by-step explanation:
Lets try to simplify g(x) since has a few extra terms:
g(x)= 3x+12-6=3x+6
Now it is easier to compare the two functions.
We can tell that they both have the same slope, both differs on a extra term
This term tell us that the g(x) is shifted to the left (it is positive 6)
Another approach to the solution is to plot the two functions together by obtaining the crossing points with the 'y' axis and with the 'x' axis
the result is shown in the attached picture
