The tangent to y=x^2e^x at x=1 cuts the x and y-axes at A and B respectively. Find the coordinates of A and B.?

y=x^2e^x

2 Answers
Jan 17, 2018

See below

Explanation:

For the slope of the tangent,

dy/dx = x^2*e^x+e^x*2(x)=x^2*e^x+2e^x x

the slope of the tangent at x=1

dy/dx = 3e

Putting x=1 in the given equation,

y = x^2*e^x

y = e

So, the tangent passes through the point (1,e)
And, it has a slope of dy/dx = 3e

So the equation of the tangent is given by,

y = mx +c

where m is the slope.

Substituting the values,

y = 3ex + c

It also passes through (1,e)

So,

e = 3e + c

c = -2e

therefore the equation of the tangent is,

y = 3ex -2e

For point A, put y=0

x=2/3

Therefore point A is,

A(2/3,0)

For point B, put x=0

y = -2e

Therefore point B is,

B(0,-2e)

Jan 17, 2018

A=(2/3,0) and B=(0,-2e)

Explanation:

We have a curve given by the equation:

y=x^2e^x

First we note that when x=1, we have:

y = 1e^1 =e

So the tangent passes through P(1,e)

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the equation we have:

dy/dx = (x^2)(d/dxe^x) + (d/dxx^2)(e^x)
\ \ \ \ \ \ = x^2e^x + 2xe^x
\ \ \ \ \ \ = (x^2 + 2x)e^x

And so the gradient of the tangent at x=1 is given by:

m = [dy/dx]_(x=1)
\ \ = (1+2)e^1
\ \ = 3e

So, using the point/slope form y-y_1=m(x-x_1) the tangent equations is;

y - e = 3e(x-1)
:. y - e = 3ex-3e)
:. y = 3ex-2e

At A we have y=0:

:. 0 = 3ex-2e => x=2/3

At B we have x=0:

:. y = -2e

Hence the required coordinates are:

A=(2/3,0) and B=(0,-2e)

(Rounding e to 2dp then P=(1,5.44) and the tangent equation is y=8.15x-5.44 making A=(0.67,0) and B=(-5.44,0))

We can verify this solution graphically:

Steve M using AutographSteve M using Autograph