Answer:
I think you forgot to add the picture.
f(x)=lnx
y=f(x)
dy/dx= 1/x
tangent at (x,y) has slope 1/x
eqn of tangent is y = mx + c
since the tangent passes through origin, c=0
substitute y = lnx and m= 1/x to above eqn
lnx = 1
x=e
y=lne=1
Answer:
y=-3/2x+6
Step-by-step explanation:
You can find the slope by taking two points on the graph, and making the one that occurs earlier in the graph (from left to right) the first point (x1, y1) and the one that occurs later in the graph the second point (x2, y2). The equation is m (or slope)=(y2-y1)/(x2-x1). I took the first two points in the table for this. m=(12-18)/-4-(-8)
the double negative on the bottom becomes an addition=> (12-18)/(-4+8)
the top simplifies to be -6 and the bottom simplifies to 4=>-6/4
this fraction can be reduced to -3/2, which is the slope of the graph.
Now, use point slope form (y-y1=m(x-x1)) to find the equation of the graph. Plug any coordinate on the graph in for x1 and y1 here. It should be correct as long as it is a point on the graph, but I am using the point (-8, 18) here.
=>y-18=-3/2(x-(-8))
the double negative in the parentheses becomes a positive=> y-18=-3/2(x+8)
distribute the -3/2 to every term in the parentheses=> y-18=-3/2x-12
add 18 to both sides, cancelling out the -18 on the left side of the equation=>y=-3/2x+6 (-12+18=6 to get 6 for b).
Therefore, the equation is y=-3/2x+6
