# Math Help - tangent to curve

1. ## tangent to curve

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?

2. At which point(s) on the curve is the slope of the curve equal not only to $-32x^{-3}$ but equal also, in particular, to ... (what value would it be interesting to find the slope equal to?) ...?

3. Originally Posted by skeske1234
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?
first find the point where the slope of the curve equal the slope of the line (i.e the slope of the curve equal 4 ) then substitute it in the curve function and the line function if the value of (y) is the same in each function then the line is tangent of the curve

4. Originally Posted by tom@ballooncalculus
At which point(s) on the curve is the slope of the curve equal not only to $-32x^{-3}$ but equal also, in particular, to ... (what value would it be interesting to find the slope equal to?) ...?

how would I find the point on the curve that has the slope of -32x^(-3) and m=4 (in the linear function)?

5. Originally Posted by Amer
first find the point where the slope of the curve equal the slope of the line (i.e the slope of the curve equal 4 ) then substitute it in the curve function and the line function if the value of (y) is the same in each function then the line is tangent of the curve
what am i doing wrong?

m=4
-32(4)^-3
y=-0.5
on the curve function

m=4

4(4)-y+11=0
27=y

??

6. Originally Posted by skeske1234
how would I find the point on the curve that has the slope of -32x^(-3) and m=4 (in the linear function)?
the slope of the line is 4 right

$y=4x+11$

$y'=4$ this is the slope

now we want the point on the curve where the slope equal 4

$y=\frac{16}{x^2}-1$

$y'=\frac{-2(16)}{x^3}=4$

$-32=4x^3$

$x^3=-8 \Rightarrow x=-2$ this is the x-coordinate for the point

now sub -2 in the function of the line

$y_{-2} = 4(-2)+11 = 3$ so the point is (-2,3)

now sub in the function of the curve

$y_{-2} = \frac{16}{(-2)^2} - 1$

$y=4-1 = 3$ the point is (-2,3) it is the same point
so the line is the tangent of the curve at the point (-2,3)

7. Originally Posted by skeske1234
what am i doing wrong?

m=4
-32(4)^-3 ???
y=-0.5
on the curve function

??
x does not equal 4, the slope = 4.

$-\frac{32}{x^3} = 4$

solve for x in the above equation, get the point on the curve, then find the tangent line equation to that point on the curve ... compare the result to the line given in the original problem statement.

8. 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 $y=16/x^2 - 1$ (i.e. $y' = -32/x^3$). 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.