Question

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`

Answer 1

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)`

Answer 2

`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: