Tangent to a Curve

[asteroidfodder]

In a section on discriminants, there is the following problem:

Prove that y = x - 3 is a tangent to the curve y = x[sup:3clsv2ya]2[/sup:3clsv2ya] - 5x + 6.

Not knowing where to begin, I hacked at it a bit, finding the solutions for y = 0:

= (x - 3) (x - 2) so x = 3 or x = 2.

Ok, so at x = 3, there is a common point (3,0).

I also found the discriminant, -1.

I thought of setting x - 3 = x[sup:3clsv2ya]2[/sup:3clsv2ya] - 5x + 6 which is (x - 3)[sup:3clsv2ya]2[/sup:3clsv2ya] but what does this imply?

None of this gets me closer to knowing that the line is actually tangent to the curve. A hint would be welcome. Thanks.

 

[soroban]

Hello, asteroidfodder!

You were on the right track (sort of).

\(\displaystyle \text{Prove that }\,y \,=\, x - 3\,\text{ is a tangent to the curve }\,y \,=\, x^2- 5x + 6\)

Two graphs are tangent if they intersect at exactly one point.


Set the two function equal to each other:

. . \(\displaystyle x^2 - 5x + 6 \:=\:x-3 \quad\Rightarrow\quad x^2 - 6x + 9 \:=\:0 \quad\Rightarrow\quad(x-3)^2 \:=\:0\)


\(\displaystyle \text{We have }one\text{ root: }\:x\,=\,3\)

\(\displaystyle \text{Hence, the curves intersect at }one\text{ point: }\3,0)\)


\(\displaystyle \text{Therefore, the line is }tangent}\text{ to the parabola.}\)

 

[arthur ohlsten]

y=x-3
slope 1
==========================
y=x^2-5x+6 slope?
dy/dx=2x-5 set dy/dx=1
1=2x-5
x=3 then y= [9-15+6] y=0
is the point x=3 y=0 on the line
y=x-3 yes
Arthur

 

[asteroidfodder]

Thanks. I got confused by the last point - I didn't understand that any straight line (except vertical), no matter how steep, would eventually "catch up" with the parabola and cross it a second time if it wasn't tangent.

Sigh, life will be so much easier when we start calculus.