SOLUTION: Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2. y = –2x + 2 y = –x + 12 y = –x + 2 y = –2x – 12

Algebra ->  Finance -> SOLUTION: Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2. y = –2x + 2 y = –x + 12 y = –x + 2 y = –2x – 12      Log On


   

Question 992501: Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2.
y = –2x + 2
y = –x + 12
y = –x + 2
y = –2x – 12
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
A tangent line can be thought of as a line that "intersects
a curve twice at the same point".

The value of y when x = 2 is

y+=+-x%5E2+%2B+3x+%2B+8
y+=+-%282%29%5E2+%2B+3%282%29+%2B+8
y+=+-4+%2B+6+%2B+8
y+=+10

So we want the equation of a tangent line at the point (2,10)

That is, we want to find the equation of the line through (2,10)
that is tangent to the parabola whose is equation is 
y+=+-x%5E2+%2B+3x+%2B+8



Let the tangent line have the equation 

y+=+mx%2Bb

The given equation of the parabola is

y+=+-x%5E2+%2B+3x+%2B+8

We set the two expressions for y equal to each other:

mx%2Bb+=+-x%5E2%2B3x%2B8
x%5E2%2Bmx-3x%2Bb-8+=+0
x%5E2%2B%28m-3%29x%2B%28b-8%29

That quadratic will have a double solution if its
discriminant is 0

The discriminant is B%5E2-4AC.  We use capital
letters to distinguish little b and capital B.

discriminant+=+B%5E2-4AC=%28m-3%29%5E2-4%281%29%28b-8%29+=+%28m-3%29%5E2-4%28b-8%29

%28m-3%29%5E2-4%28b-8%29=0

Solve that for b

m%5E2-6m%2B9-4b%2B32=0
m%5E2-6m-4b%2B41=0
m%5E2-6m%2B41=4b
expr%281%2F4%29%28m%5E2-6m%2B41%29=b

Substitute that for b in

y+=+mx%2Bb

y+=+mx%2Bexpr%281%2F4%29%28m%5E2-6m%2B41%29

And also the point (x,y) = (2,10)

10+=+m%282%29%2Bexpr%281%2F4%29%28m%5E2-6m%2B41%29

10+=+2m%2Bexpr%281%2F4%29%28m%5E2-6m%2B41%29

Multiply through by 4 to clear fraction:

40+=+8m%2B%28m%5E2-6m%2B41%29

40+=+8m%2Bm%5E2-6m%2B41

0+=+m%5E2%2B2m%2B1%29

0+=+%28m%2B1%29%5E2

m%2B1=0

m=-1

Substitute this and (x,y) = (2,10) in

y+=+mx%2Bb

10+=+%28-1%29%282%29%2Bb

10+=+-2%2Bb

12=b

So the equation of the line is 

y+=+-x%2B12

So we draw that line on the graph to see if it looks like what we
have calculated. It has y-intercept and slope -1, and goes through
points (-1,13), (0,12), (4,8)



Edwin