Where functions meet

[Nikolas111]

Could somebody help me with this trouble .Find the values of m for which the lines y=mx-2 are tangents to the curve with equation y=x^2-4x+2 I always get m=0 1. mx-2=x^2-4x+2
2. x^2+(-4-m)x+4
3.b^2-4a*c
4.16+m^2-16=0
5.m^2=0
6.m=0
The trouble is that the answer is that m= 0 or 8 and I don't know how I should count it to get both answers .


Thank you in advance.

 

[stapel]

Find the values of m for which the lines y = mx - 2 are tangents to the curve with equation y = x² - 4x + 2.

I always get m = 0. My steps are:

1. mx - 2 = x^2 - 4x + 2
2. x^2 + (-4-m)x + 4

How did you get this? If this came from line (1), what happened to the "equals" sign and whatever was on the other side?

3. b^2-4a*c

What is this? Where did it come from? What was your reasoning for using this?

4. 16+m^2-16=0

Where did this come from? What happened to "x"? Are you saying that your reasoning and steps were something similar to the following?

. . .\(\displaystyle \mbox{I want to find the x-values where }\,y_1\, =\, mx\, -\, 2\, \mbox{ and }\)

. . .\(\displaystyle y_2\, =\, x^2\, -\, 4x\, +\, 2\, \mbox{ intersect with the same sl}\mbox{ope.}\)

If so, then how were you to account for the "slope" of y₂, which is curved?

. . .\(\displaystyle \mbox{I'll solve the quad}\mbox{ratic created by setting }\, y_1\, \mbox{ equal to }\, y_2:\)

. . . . .\(\displaystyle mx\, -\, 2\, =\, x^2\, -\, 4x\, +\, 2\)

. . . . .\(\displaystyle 0\, =\, x^2\, -\, 4x\, -\, mx\, +\, 2\, +\, 2\)

. . . . .\(\displaystyle 0\, =\, x^2\, +\, \left(-4\, -\, m\right)\, x\, +\, 4\)

. . .\(\displaystyle \mbox{Then }\, a\, =\, 1,\, b\, =\, -4\, -\, m,\, \mbox{ and }\, c\, =\, 4.\)

These variables, "a", "b", and "c, I'm guessing came from the Quadratic Formula, which you then applied:

. . . . .\(\displaystyle \begin{align} x\, &=\, \dfrac{-(-4\, -\, m)\, \pm\, \sqrt{\strut (-4\, -\, m)^2\, -\, 4(1)(4)\,}}{2(1)}

\\ \\ &=\, \dfrac{4\, +\, m\, \pm\, \sqrt{\strut (16\, +\, 8m\, +\, m^2)\, -\, 16\,}}{2}

\\ \\ &=\, \dfrac{4\, +\, m\, \pm\, \sqrt{\strut m^2\, +\, 8m\,}}{2} \end{align}\)

How did you arrive at your later steps? How does any of the above take into account the need for y₁ to be tangent to the parabola? Since you're in algebra (rather than calculus), how are you supposed to find the slope of the parabola?

Please be complete. Thank you!

 

[Nikolas111]

When functions meet

Thank you very much :wink:

I didn't know how to solve that problem so therefore I made those mistakes .

 

[stapel]

I didn't know how to solve that problem....

One usually uses calculus. Since you're in an algebra class (before calculus), your class must have been given some other method. We won't be able to help you with whatever that method was until you provide us with that information. Please be complete. Thank you!