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
Yes, you have done the problems correctly! Great job!
To find the equation of the tangent line at a given value, you followed the correct steps:
1. First, you found the derivative of the function to get the slope of the tangent line.
2. Then, you plugged in the given value into the original equation to find the corresponding y-value.
3. You used the point-slope formula, y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point on the curve, to write the equation of the tangent line.
For problem a), you correctly found the slope to be 1 and the y-value to be 0 when x = -3. The equation of the tangent line is y - 0 = 1(x - (-3)), which simplifies to y = x + 3.
For problem b), you correctly found the slope to be -1/2 and the y-value to be 3 when x = 1. The equation of the tangent line is y - 3 = -1/2(x - 1).
Both of these equations can be simplified further, as you mentioned. The equation for problem a) simplifies to y = x + 3, and the equation for problem b) simplifies to y = -1/2x + 5/2.
Overall, your solutions are correct, and simplifying the equations further is a good idea to make them more concise. Keep up the good work!