At which point(s) on the curve is the slope of the curve equal not only to but equal also, in particular, to ... (what value would it be interesting to find the slope equal to?) ...?
Show that the line 4x-y+11=0 is tangent to the curve y=16/x^2 -1
What I have so far:
I found out that the slope of the linear function is m=4
Then I found the derivative of the curve which ended up to be
y'=-32x^(-3)
so now what should I do with this to show that the line 4x-y+11=0 is tangent to the curve?
the slope of the line is 4 right
this is the slope
now we want the point on the curve where the slope equal 4
this is the x-coordinate for the point
now sub -2 in the function of the line
so the point is (-2,3)
now sub in the function of the curve
the point is (-2,3) it is the same point
so the line is the tangent of the curve at the point (-2,3)
You could also approach this problem by finding where the 2 functions intersect by setting them equal. After you find the x values where they are equal, then you can substitute those x values into the slope function of (i.e. ). Then if the resulting slope = 4, you know the lines are tangent. Hopefully this wasn't too confusing, but it ensures that you do in fact check if the lines even intersect, before assuming they do.