Question

Find the slope of the tempest on the eve y=x^2+3x+4 (-1,2)
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Answer 1

Step-by-step explanation:

Line passing through origin : y = mx, slope of tangent of a curve is

dx

dy

at that point.

∴y=x

2

+3x+4

⇒

dx

dy

=2x+3.

Pt.(x,y)

y

1

=mx

1

→(i)

y

2

=x

2

1

+3x

1

+4→(ii)

m=2x

1

+3

Putting value in (i)

y

1

=2x

1

2

+3x

1

→(iii)

Solve (ii) & (iii)

2x

2

1

+3x

1

=x

1

2

+3x

1

+4

x

1

2

=4

x

1

=±2

x

1

=2,y

1

=4+6+4=14

x

1

=−2,y

1

=(−2)

2

−3×2+4=4−6+4=2

∴ Points an (2,14) and (−2,2)

Answer 2

Answer:

Let the point is P(a,b)

Now the given curve is y=x

2

+3x+4

Differentiating w.r.t x

dx

dy

=2x+3

Thus slope of tangent at P is =(

dx

dy

)

(a,b)

=2a+3

Hence equation of tangent at P is given by, (y−b)=(2a+3)(x−a)

But given that tangent passes through origin

⇒(0−b)=(2a+3)(0−a)⇒2a

2

+3a=b..(1)

Also the point P lies on the given curve,

b=a

2

+3a+4...(2)

Solving (1) and (2) we get the required point P coordinate

which are (−2,2)or(2,14)

Hence, option 'A' is correct.