Finding a tangent line ASAP!

[thomcart8]

Ok i needed to find a tangent line for some constant b and x > 0, let y = xlnx-bx. I needed to find the equation of the tangent line at the point where the graph crosses the x axis.

Ok i came up with y = xlnx-((xlnx)/x)*x). Is this right or even close?

 

[Deleted member 4993]

Ok i needed to find a tangent line for some constant b and x > 0, let y = xlnx-bx. I needed to find the equation of the tangent line at the point where the graph crosses the x axis.

Ok i came up with y = xlnx-((xlnx)/x)*x). Is this right or even close?

The expression you wrote turns into y = 0

If you want us to check - you must post your work.

 

[iambryanatkinson]

to find equation of tangent line there is two things you have to find : slope and intercept

the 1st derivertive at the point will give you slope

put the slope in a linear equation y = mx + c ; which will be the equation of the tangent line

since tangent line and the equation in the problem are intersected at tangent point,
we can solve two equation to get the intercept then you get complete equation of tangent line

 

[Deleted member 4993]

Ok i needed to find a tangent line for some constant b and x > 0, let y = xlnx-bx. I needed to find the equation of the tangent line at the point where the graph crosses the x axis.

Ok i came up with y = xlnx-((xlnx)/x)*x). Is this right or even close?

Point of x-intercept y= 0 ? x = e[sup:3awyy6gc]b[/sup:3awyy6gc]

slope of the tangent line at that point

dy/dx = ln(x) + 1 - b

at the point of x-intercept (x = e[sup:3awyy6gc]b[/sup:3awyy6gc]) slope of the tangent line m = 1

At the point of x-intercept, equation of the tangent line

(y - 0) = 1 (x - e[sup:3awyy6gc]b[/sup:3awyy6gc])

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