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
See below
Explanation:
For the slope of the tangent,
the slope of the tangent at
Putting
So, the tangent passes through the point (1,e)
And, it has a slope of
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,
For point A, put
Therefore point A is,
For point B, put
Therefore point B is,
# 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
# 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: