Question

Find the value(s) of k such that the graphs of the quadratic functions f(x)=(x+3)^2+k and g(x)=-(x+k)^2+3 are tangent to one another.

Answer 1

is this calculus?

Answer 2

No this should be solved using Algebra 2

Answer 3

I've tried making the equations equal to each other and solving for k by setting the discriminant equal to 0 but my answers were wrong. The correct answer is 1 and 3 but I need to know how to get there

Answer 4

Press play and you will see those answers.
https://www.desmos.com/calculator/alw9wn9kk2

I am not really sure how to do this without calculus (i'll look into the problem)

Answer 5

I know how they arrive at k=3 with algebra

Answer 6

3 is easy but I don't know how to get 1

Answer 7

So, you know how to get k=3?

Answer 8

Yes

Answer 9

\(\color{#000000 }{ \displaystyle y=(x+3)^2 }\)
\(\color{#000000 }{ \displaystyle y=-(x+k)^2 }\)

They are tangent, at k=3, because then they are reflected about x=0, and we know that in that case k=3, so if you say

\(\color{#000000 }{ \displaystyle y=(x+3)^2 +k }\)
\(\color{#000000 }{ \displaystyle y=-(x+k)^2 +3 }\)

then we use the same fact, but this time they are reflected about x=3.

Answer 10

i think i have seen a problem like this before, but i kind of forget how to do it
i think it has something to do with setting them equal, then making sure they intersect at only one point by making sure the discriminant is zero

Answer 11

we can try it and see if we get the two answers you know to be true

Answer 12

I tried that but the equation for k didn't give correct values. I'll try again though

Answer 13

did you start with \[-(x+k)^2+3=(x+3)^2+k\]?

Answer 14

i need paper to do this, hold the phone

Answer 15

I tried it again and got the correct answers. I must've just done some calculation error on the first attempt sorry

Answer 16

i did too, let me try again

Answer 17

ok not the second time either damn

Answer 18

third time is a charm! got it

Answer 19

start here
\[-(x+k)^2+3=(x+3)^2+k\]

Answer 20

expand and set equal to zero
you want my work?

Answer 21

@SolomonZelman i have seen this before, to do without calc
think i googled it or something
ideas is to set them equal, then take the discriminant, set it equal to zero and solve, thereby making sure there is only one solution to the intersection i.e. they are tangent

Answer 22

confused the hell out of me the first time i saw it

Answer 23

@WolframWizard you get it yet or no?

Answer 24

Yes, satelitte, when you said this thing about setting the discriminant zero I figured it out ... just that not everytime I can come up with something; because figuring after you've seen someone is easier. tnx for the information!

Answer 25

turns out in this case if you set \(b^2-4ac=0\) you get \(k=3,k=1\)
never thought of your other reason before
wonder if you can come up with the 1 by sheer reason as well?

Answer 26

well I wil expect that calculus is sheer.

Answer 27

lol

Answer 28

But, not for this post since the user needs to solve with algebra...

Answer 29

perhaps there is some other way of solving it besides calc and besides \(\Delta =0\)