For each problem, find the equation of the tangent line
to the function at the given point.
Problem 1 :
f(x) = x2 + 1; (1, 2)
A) y = 2x C) y = (1/2)x + 3/2 |
B) y = -8x + 10 D) y = -6x + 8 |
Problem 2 :
f(x) = 2x2 + 2x + 2; (-1, 2)
A) y = -8x – 6 C) y= (1/2)x + 5/2 |
B) y = -2x D) y = 6x + 8 |
Problem 3 :
y = x2 + x - 2; (1, 0)
C) y = 3x - 3 |
D) y = -12x + 12 |
Problem 4 :
y = 2x2 + 1; (-1, 3)
A) y = 2x + 5 B) y = -4x – 1 C) y = 12x + 15
Problem 5 :
A) y = (1/4)x + 1 C) y = -x + 1 |
B) -2x + 1 D) y = 1 |
Problem 6 :
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Problem 7 :
A) y = -1 C) y = -(1/2)x - 2 |
B) y = x + 1 D) y = 4x + 7 |
Problem 8 :
1) y = 2x, option (A)
2) y = -2x, option B
3) y = 3x - 3, option C
4) y = -4x - 1, option B
5) y = -x + 1, option C
6) y = (1/4)x - 1, option C
7) y = x + 1, option B
8) y = -(1/4) x - (3/4)
For each problem, find the equation of the line tangent to the function at the given point. Your answer should be in slope – intercept form.
Problem 1 :
y = x3 – 3x2 + 2 at (3, 2)
Problem 2 :
Problem 3 :
y = x3 – 2x2 + 2 at (2, 2)
Problem 4 :
Problem 5 :
Problem 6 :
Problem 7 :
y = In (-x) at (-2, In 2)
Problem 8 :
y = -2tan (x) at (-π, 0)
1) y = 9x - 25
2) y = (-5/2)x -5
3) y = 4x - 6
4) y = -8x/27 - (23/27)
5) y = (2/3)x + 1/3
6) y = (1/2)x + 3
7) y = (-1/2)x + ln 2 - 1
8) y = -2x - 2π
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM