# 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#.?

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#y=x^2e^x#

See below

For the slope of the tangent,

So the equation of the tangent is given by,

where m is the slope.

Substituting the values,

It also passes through (1,e)

So,

therefore the equation of the tangent is,

Therefore point A is,

Therefore point B is,

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

We have a curve given by the equation:

# y=x^2e^x #

First we note that when

# y = 1e^1 =e #

So the tangent passes through

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

# m = [dy/dx]_(x=1) #

# \ \ = (1+2)e^1 #

# \ \ = 3e #

So, using the point/slope form

# y - e = 3e(x-1) #

# :. y - e = 3ex-3e) #

# :. y = 3ex-2e #

At

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

At

# :. y = -2e #

Hence the required coordinates are:

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

(Rounding

We can verify this solution graphically:

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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