Question

Find an equation of the tangent line to the graph of P(x)=square root of (H (x)) when x=-4

Answer 1

sorry I have to attach a file because I am unable to draw the given graph with my mouse

Answer 2

I used the chain rule and sub x=0 but I got (1/2)times (1/square root of 0)
not sure I did this correctly at all

Answer 3

ok so we have \(P(x)=\sqrt{H(x)}=(H(X))^{\frac{1}{2}}\)
so
\(P'(x)=\frac{1}{2}*(H(x))^{-\frac{1}{2}}H'(x)=\frac{H'(x)}{2\sqrt{H(x)}}\)
\(H'(-4)=-\frac{1}{2}\) and \(H(-4) = 4\)
so
\(P'(-4) =\frac{-\frac{1}{2}}{2\sqrt{4}}=-\frac{1}{8}\) this right so far?

Answer 4

omg I use x=4 not x=-4

Answer 5

:)

Answer 6

Let me go back and redo it with x=-4

Answer 7

Thanks, boy I can't believe I used x=4 and not x=-4

Answer 8

I really like @zzr0ck3r 's solution, clever stuff!

Answer 9

It's just the slope of the line at that location. If you look at the graph it looks like this: And if you think of slope as being rise over run, then it rises negative 1 and runs 2.

Answer 10

look at the graph, slope of straight line

Answer 11

whats the problem? we are not one....that will just give the slope of the tangent line at that point.

Answer 12

we are not done*

Answer 13

but the required tangent line is not for H(x), it is P(x), right?

Answer 14

correct, and I found P'(-4)

Answer 15

this gives slope.

Answer 16

ok so the tangent line is just y-4=(-1/8)(x-4)

Answer 17

the function of P(x) is P(x) = \(\sqrt{H(x)}=\sqrt{\dfrac{x}{2}+2}\) for the part contains x =-4 of H(x) and it has the form of

Answer 18

yes we did not find the tangent line

Answer 19

I think your graph has problems.

Answer 20

to find the tangent line of P(x) at a, we get slope by P'(a). this is what I did.

Answer 21

The tangent line, which let's call it f(x) why not, currently looks like this:

f(x)=-x/8+B

We still need to solve for the intercept of our tangent line, and in order to do this we use the fact that P(x)=f(x) when x=-4.

Answer 22

now we have \(y=-\frac{1}{8}x+b\) and need to find \(b\). So we use the point \((-4,P(-4))=(-4,2)\) and get \(2=-\frac{1}{8}(-4)+b\implies b = 4\) So our line is \(y=-\frac{1}{8}x+4\)

Answer 23

Thanks for explanation, since I think we have to have the function P(x) first which defined by sqrt (H). This graph is totally different from the given graph. But if you guys said so. hihihi... I am ok.

Answer 24

is my tangent line incorrect?

Answer 25

yes, this line y I define above is not tangent to any point on the graph shown, it is tangent to the point on P(x) which is not shown

Answer 26

@precal I am sorry for messing up your post.

Answer 27

no problem, I am all about doing problems with the goal of understanding them