Application of DerivativeHard
Question
The slope of the tangents to the curve y = (x + 1) (x- 3) at the points where it crosses x- axis are-
Options
A.± 2
B.± 3
C.± 4
D.None of these
Solution
y = (x + 1) (x - 3)
Where it crosses to x-axis
put y = 0
0 = (x + 1) (x - 3)
x = -1, 3
(-1 , 0) & (3, 0)
= (x - 3) + (x + 1) = 2x -2
= 2(x - 1) ⇒ 
= - 4
= 4 = ± 4
Where it crosses to x-axis
put y = 0
0 = (x + 1) (x - 3)
x = -1, 3
(-1 , 0) & (3, 0)
= (x - 3) + (x + 1) = 2x -2 = 2(x - 1) ⇒ = - 4 = 4 = ± 4Create a free account to view solution
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