Finding the equation of a tangent without using calculus

saroo2a

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Jun 20, 2014
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Hi,
If I was told to find the equation of line with slope (1) that is tangent to the curve of a rational function, 1/-x-3 , would I get two tangent lines at the end?
Important :please do not use calculus or derivatives.
Here are the steps I use...
I first find the general equation of the line, which would be y=(1)x+b
then i let the rational function = the linear function
Then I rearrange the equation so that everything is on the left side and zero is on the left side.
Then, I make the equation be in the standard form of a quadratic,
Then I use the discriminant and let it equal zero, (previous notes that i have say :so that we may have one solution to the quadratic,one double root)
Then solve for b
However at the end My answer looks like this:
(b-5)(b-1)=0
so b=5 or b=1
so that way i would have two equations:
y=x+5 and y=x+1
Although it makes perfect sense that they are tangents ,however, doesnt this mean that i have two solutions ?
Thank you :)
 
Hi,
If I was told to find the equation of line with slope (1) that is tangent to the curve of a rational function, 1/-x-3 , would I get two tangent lines at the end?
Important :please do not use calculus or derivatives.
Here are the steps I use...
I first find the general equation of the line, which would be y=(1)x+b
then i let the rational function = the linear function
Then I rearrange the equation so that everything is on the left side and zero is on the left side.
Then, I make the equation be in the standard form of a quadratic,
Then I use the discriminant and let it equal zero, (previous notes that i have say :so that we may have one solution to the quadratic,one double root)
Then solve for b
However at the end My answer looks like this:
(b-5)(b-1)=0
so b=5 or b=1
so that way i would have two equations:
y=x+5 and y=x+1
Although it makes perfect sense that they are tangents ,however, doesnt this mean that i have two solutions ?
Thank you :)

As you wrote it. the rational function looks like

\(\displaystyle y = -\dfrac{1}{x} - 3\) .......... Is that the correct function?

Or did you mean to write:

\(\displaystyle y = -\dfrac{1}{x + 3} \)
 
As you wrote it. the rational function looks like

\(\displaystyle y = -\dfrac{1}{x} - 3\) .......... Is that the correct function?

Or did you mean to write:

\(\displaystyle y = -\dfrac{1}{x + 3} \)
Hi,
The function is 1/(-x-3)
In words that is : one over negative x minus 3, where negative x minus 3 is in the denominator.
I hope thats a good description
Thank you
 
Hi,
If I was told to find the equation of line with slope (1) that is tangent to the curve of a rational function, 1/(-x-3) ,

Those parentheses are important for correct representation of the function

would I get two tangent lines at the end? Yes

Important :please do not use calculus or derivatives.

Here are the steps I use...
I first find the general equation of the line, which would be y=(1)x+b
then i let the rational function = the linear function
Then I rearrange the equation so that everything is on the left side and zero is on the left side.
Then, I make the equation be in the standard form of a quadratic,
Then I use the discriminant and let it equal zero, (previous notes that i have say :so that we may have one solution to the quadratic,one double root)
Then solve for b
However at the end My answer looks like this:
(b-5)(b-1)=0
so b=5 or b=1
so that way i would have two equations:
y=x+5 and y=x+1
Although it makes perfect sense that they are tangents ,however, doesn't this mean that i have two solutions ?
Thank you :)
There are two tangents possible with m = 1

If you plot the function within the domain [-6,1] you will see how that is possible!
 
There are two tangents possible with m = 1

If you plot the function within the domain [-6,1] you will see how that is possible!

Thanks for confirming that there are two tangents.
can you please answer my last question? what is a double root?
 
Thanks for confirming that there are two tangents.
can you please answer my last question? what is a double root?

Quadratic equations have two roots. When those two roots are congruent (discriminant = 0), then we get one "repeated root" or "double root".
 
Quadratic equations have two roots. When those two roots are congruent (discriminant = 0), then we get one "repeated root" or "double root".
So, in this case , why didnt i get one repeated root? I got two different roots, even though i set the discriminant =0 ?
Thank you
 
So, in this case , why didnt i get one repeated root? I got two different roots, even though i set the discriminant =0 ?
Thank you

You got repeated root for your original quadratic equation (x2 + x(b+3) +3b +1 =0).

You got "two" values of 'b' which satisfied the condition.

Thus (for b = 5)

x2 + x(b+3) +3b +1 = 0 → x2 + 8x +16 = 0 → (x + 4)2 = 0 → x = -4 (one repeated root)

and (for b = 1)

x2 + x(b+3) +3b +1 = 0 → x2 + 4x +4 = 0 → (x + 2)2 = 0 → x = -2 (one repeated root)

Notice by using different 'b' - you got two different quadratic equations [(x2 + 8x +16 = 0) & (x2 + 4x +4 = 0)] and two different repeated roots
 
So, in this case , why didnt i get one repeated root? I got two different roots, even though i set the discriminant =0 ?

Hi saroo2a:

You seem to be confusing symbols.

The root of the quadratic equation is a value of x; the two values 1 and 5 that you're talking about are values of b.

Here is your quadratic polynomial, right? x^2 + (b+3)x + (3b+1)

You found that symbol b can represent either 1 or 5.

Substituting b=1 into this polynomial gives x^2 + 4x + 4

Substituting b=5 gives x^2 + 8x + 16

For each of these polynomials, the discriminant is zero and the root is repeated.

Cheers :)
 
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