If the tangent to the curve y = x3+ax+b at (1,-6) is perpendicular to the line 2x+2y+7=0, find a and b
Answers
Answer:
Given: y=x
3
+ax+b ........(i)
Tangent to the curve=
dx
dy
=3x
2
+a
Tangent to the curve at point (1,−6) =
dx
dy
(1,−6)
=3(1)
2
+a=3+a
Now, x−y+5=0
⇒y=x+5
Comparing it with y=mx+c, where m= slope, we get m=1
Therefore, slope of the above line is 1.
Given tangent is parallel to the line x−y+5=0.
Thus slope will be equal.
Therefore, 3+a=1
⇒a=1−3=−2
Substituting a=−2 in eqn (i) we get
−6=1
3
+(−2)(1)+b
⇒−6=1−2+b
⇒−6+1=b
⇒b=−5
⇒a=−2,b=−5
Step-by-step explanation:
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Answer:
a=-2,b=-5
Step-by-step explanation:
Tangent to the curve= dxdy=3x2+a
Tangent to the curve at point (1,−6) = dxdy(1,−6)=3(1)2+a=3+a
Now, x−y+5=0
⇒y=x+5
Comparing it with y=mx+c, where m= slope, we get m=1
Therefore, slope of the above line is 1.
Given tangent is parallel to the line x−y+5=0.
Thus slope will be equal.
Therefore, 3+a=1
⇒a=1−3=−2
Substituting a=−2 in eqn (i) we get
−6=13+(−2)(1)+b
⇒−6=1−2+b
⇒−6+1=b
⇒b=−5
⇒a=−2,b=−5