Question

For which values of m and n does the graph of f(x)=mx+n intersect the graph of g(x)=1/x in exactly one point and also contain the point (-1,1)?

Answer 1

f(x)=mx+n has the point (-1,1) on it when f(-1) = 1


f(x)=mx+n
f(-1)=m(-1)+n ... replace x with -1
f(-1)=-m+n
1=-m+n ... replace f(-1) with 1
m=n


making sense so far?

Answer 2

Yes

Answer 3

I wouldn't use point slope form would I?

Answer 4

not really

Answer 5

I guess you could, but I don't see how it comes into play

Answer 6

Wait, why did you replace f(-1) with 1?

Answer 7

because f(-1) = 1

Answer 8

the point (-1,1) lies on f(x)

Answer 9

Okay I get up to the 1= -m+n. But how didi you get rid of the 1?

Answer 10

oh my bad, typo

Answer 11

1=-m+n
1+m=n
n = m+1

Answer 12

ready for the next part?

Answer 13

Yes.

Answer 14

We're looking for intersection points between f and g. So we set the two functions equal to each other and solve for x. In that process, we'll need to get everything to one side

f(x) = g(x)
mx+n = 1/x
x*(mx+n) = 1
mx^2+nx = 1
mx^2+nx-1 = 0

Answer 15

Oh seriously. I thought we knew what the y intercept was so we could just plug it into the equation and use trial and error

Answer 16

no we don't know the y-intercept of f(x)

Answer 17

we want to force f and g to have exactly one intersection point

that means we want to force mx^2+nx-1 = 0 to have exactly one solution

Answer 18

Okay, so would I factor? but the variables are different though

Answer 19

The equation mx^2+nx-1 = 0 has exactly one solution when the discriminant D = b^2 - 4ac is equal to zero (ie D = 0)


In this case, a = m, b = n, c = -1


D = b^2 - 4ac
0 = b^2 - 4ac ... plug in D = 0
0 = n^2 - 4m*(-1) ... plug in a = m, b = n, c = -1
0 = n^2 + 4m
0 = (m+1)^2 + 4m ... plug in n = m+1

Answer 20

at this point, you need to solve 0 = (m+1)^2 + 4m for m. You'll get 2 solutions

Answer 21

would using the quadratic formula work too. I'm not used to the discriminant

Answer 22

the discriminant is part of the quadratic formula

Answer 23

the b^2 - 4ac is under the square root

Answer 24

While solving that for m, I foiled the (m+1)^2 and after adding it to 4m, got m^2+6m+1 I'm not sure that's factorable

Answer 25

you'll have to use the quadratic formula

Answer 26

to solve m^2+6m+1 = 0

Answer 27

Okay, I know how do to it from there. But why did you plug in n=m+1 for that step
0 = n^2 + 4m
0 = (m+1)^2 + 4m ... plug in n = m+1.

Answer 28

Oh to get like terms, gotcha

Answer 29

yeah I wanted one variable to solve for