SOLUTION: Find an equation of the line tangent to each curve through the given point 1. y=x2-3x+7 (2,5)

Algebra ->  Linear-equations -> SOLUTION: Find an equation of the line tangent to each curve through the given point 1. y=x2-3x+7 (2,5)      Log On


   

Question 1071734: Find an equation of the line tangent to each curve through the given point
1. y=x2-3x+7 (2,5)
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find an equation of the line tangent to each curve through the given point
1. y=x2-3x+7 (2,5)
--------------
Use ^ (Shift 6) for exponents.
It's a bad idea to include the problem #, 1. in this case.
Sometimes it is misconstrued as part of the problem.
---------------------
y=x^2-3x+7 (2,5)
y' = 2x - 3
------------
At x = 2, y' = 1, the slope of the tangent line.
---
--> find an eqn of a line thru (2,5) with a slope of 1
-------
y - 5 = 1*(x - 2)
And we're done.


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor assumed you are taking calculus, not 
just pre-calculus or college algebra.  I will use
only the mothods taught in algebra and pre-calc in 
case you aren't taking calculus.  It's easier with
calculus, but can be done with algebra only.

The point-slope form of a line is

y-y%5B1%5D=m%28x-x%5B1%5D%29

If the line goes through (2,5) we can substitute 
(x1,y1)=(2,5) and we have:

y-5=m%28x-2%29

y-5=mx-2m

y=mx-2m%2B5

We substitute mx-2m+5 for y in

mx-2m%2B5+=+x%5E2-3x%2B7

mx-2m+=+x%5E2-3x%2B2

0=x%5E2-mx-3x%2B2m%2B2

x%5E2-%28m%2B3%29x%2B2%28m%2B1%29=0

In order for the line to be tangent,
the discriminant must = 0

b%5E2-4%2Aa%2Ac=0
%28-%28m%2B3%29%29%5E2-4%2A%281%29%2A2%28m%2B1%29=0
%28m%2B3%29%5E2-8%28m%2B1%29=0
m%5E2%2B6m%2B9-8m-8=0
m%5E2-2m%2B1=0
%28m-1%29%28m-1%29=0
%28m-1%29%5E2=0
m-1=0
m=1

So the line's equation

y=mx-2m%2B5

is now

y=1x-2%2A1%2B5

y=x-2%2B5

y=x%2B3



Edwin