MedVision ad

Integration help please (2 Viewers)

peterkim95

New Member
Joined
May 27, 2006
Messages
15
Gender
Undisclosed
HSC
N/A
Ive got an exam tomorrow, but i dunt know how you do this question :( :mad1:
prove that the line y = x + 2 is a tangent to the parabola y= x^2 - 5x + 11
Thanks for your help!
 
Joined
Mar 3, 2005
Messages
2,359
Location
Wollongong
Gender
Male
HSC
2006
solve them simultaneously...well at least combine the two equations and make it one...

y = x + 2
y= x^2 - 5x + 11

x + 2 = x^2 - 5x + 11
x^2-6x-9 = 0

discriminant of x^2-6x+9

= b^2-4ac
=36-4(1)(9)
=0

since the discriminant = 0, y = x+2 only intersects the parabola at ONE point. if the discriminant were negative it wouldn't touch the parabola at all and it were positive it would touch at two points.

therefore it must be a tangent.

hope that helps :wave:
 
Last edited:

ThuanSUX

New Member
Joined
Aug 21, 2006
Messages
9
Gender
Male
HSC
2000
An alternative solution to what Watatank said would be:

1. Combine the equations
2. Factorize

y = x+2
y = x^2-5x+11

x^2-5x+11 = x+2
x^2-6x+9 = 0
(x-3)(x-3) = 0
(x-3)^2 = 0
thus, x=3 ONLY

Resubstituting it into the first equation to obtain a y-coordinate and you get the point (3,5). Hence, the two equations only intersect at one point, making y=x+2 a tangent of y=x^2-5x+11.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top