Asked by Tammy
First I would find the first derivative which is the slope and then would I
plug in the given value into the original equation to find y and then use
y-y_1=m(x-x_1)?
Find the euation of the tangent line at the given value
a) y=ln abs(x+4) x = -3
b) y= (x^2 +5)/(x+1) x=1
a) I found the slope to be 1 and then I got the y-value as ln 1 which
equals 0. so would the equation be y-0= 1(x- -3)?
b) I got the slope as -1/2 and the y-value as 3 so then i got y-3= -1/2
(x-1)
Did I do these problems correctly?
both correct!, good job
are you going to simplify your equations a bit more?
e.g. the first reduces to y=x+3
plug in the given value into the original equation to find y and then use
y-y_1=m(x-x_1)?
Find the euation of the tangent line at the given value
a) y=ln abs(x+4) x = -3
b) y= (x^2 +5)/(x+1) x=1
a) I found the slope to be 1 and then I got the y-value as ln 1 which
equals 0. so would the equation be y-0= 1(x- -3)?
b) I got the slope as -1/2 and the y-value as 3 so then i got y-3= -1/2
(x-1)
Did I do these problems correctly?
both correct!, good job
are you going to simplify your equations a bit more?
e.g. the first reduces to y=x+3
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