Question

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​

Answer 1

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:

It is hopefull for you

Answer 2

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