Determine the equations of both lines that are tangent to the graph of f(x) = x^2 and pass through the point (1,-3).

- anonymous

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- myininaya

f'(x)=2x
f'(a)=2a
so the equation of the line is y=2ax+b
we know a point on this line (1,-3) so we
-3=2a(1)+b
-3=2a+b
....

- myininaya

thinking...

- anonymous

One would have a negative slope, the other would be positive

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## More answers

- myininaya

right

- anonymous

Would those two lines be orthogonal to each other?

- myininaya

that looks possible it looks like it could form a 90 degree angles

- anonymous

Ok so then they must have reciprocal slopes

- myininaya

opposite reciprocal slopes

- myininaya

wait i htink i'm fixing to figure something out....

- myininaya

##### 1 Attachment

- myininaya

now we need to figure the y intercept for each line which is easy since we know a point on both lines. both lines have the point (1,-3)

- myininaya

y=6x+b (1,-3)
-3=6(1)+b
-3=6+b
-3-6=b
-9=b
so y=6x-9 is the line that is tangent to the point (3,9) on the curver y=x^2 that passes through (1,-3)
y=-2x+b (1,-3)
-3=-2(1)+b
-3+2=b
-1=b
so y=-2x-1 is the line tangent to the point (-1,1) on the curve y=x^2 that passes through (1,-3)

- myininaya

we can check if you want
Let's find the tangent line at (3,9) and see if passes through (1,-3)
so f'(x)=2x
f'(3)=6
y=6x+b
9=6(3)+b
9-18=b
b=-9
so y=6x-9
is (1,-3) on that line
-3=6(1)-9=-3 so yes!
now let's check the tangent line at (-1,1) and see if passes through (1,-3)
so f'(x)=2x
f'(-1)=-2
y=-2x+b
1=2+b
b=-1
y=-2x+-1
is (1,-3) on that line
-3=-2(1)+-1=-3 so yes!
:) we are finished!

- myininaya

so the lines arent orthogonal

- myininaya

do you have any questions?

- anonymous

One question

- anonymous

Why did you choose the general points (a,a^2) to be X1,Y1 respectively? Does it make a difference if I were to use them as X2,Y2 in the slope formula?

- myininaya

(a^2-(-3))/(a-1)=(-3-a^2)/(1-a)
we can make both sides look the same
(-1)/(-1)=1
so if we multiply our second fraction by (-1)/(-1) we would get the first fraction or vice versa

- myininaya

(y1-y2)/(x1-x2)=(y2-y1)/(x2-x1)
(-1)/(-1)*(y1-y2)/(x1-x2)=(y2-y1)/(x2-x1)

- myininaya

a-b is the same as -(b-a)

- myininaya

(6-2)/(3-1)=4/2=2
(2-6)/(1-3)=-4/(-2)=2

- myininaya

cool?

- anonymous

Yes thank you very much. I have just a few more questions if you don't mind helping me out with them.

- myininaya

go for it

- anonymous

If f(a) = 0 and f'(a) = 6, find the limit h --> 0 of:
f(a + h)
-------
2h

- myininaya

i got it scanning it now

- myininaya

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- myininaya

i just used the definition of derivative

- myininaya

and plugged and what they gave me and solved for limh->0 (f(a+h)/(2a))

- myininaya

do you understand both problems we did?

- myininaya

oops solved for limh->0(f(a+h)/(2h))

- anonymous

I don't understand how you did the second question.

- myininaya

what is definition of the derivative at a of f

- anonymous

Wait, ok wouldn't it be 12 for the answer because:
6 = f(a+h)/h * (1/2)
6(2) = f(a+h)/h

- anonymous

Sorry I don't know what happened

- myininaya

doesn't the question above ask for lim h->0 (f(a+h)/(2a))

- myininaya

oops the bottom is suppose to read 2h

- myininaya

so we have 6=limh->0(f(a+h)/(2h))

- myininaya

oops 6=limh->0(f(a+h)/h) do you understand this far?

- anonymous

yes

- myininaya

ok but the want 1/2* limh->0(f(a+h)/h)
so if we multiply the above equation on one side by 1/2
don't we have to do it to the other side?

- anonymous

yes

- myininaya

ok 6*(1/2)=3

- anonymous

Got it. Thank you.

- anonymous

Ok last question for now is this: Determine the value of "a', given that the line "ax - 4y + 21 = 0" is tangent to the graph y = a/x^2 at x = -2

- myininaya

if you write that tangent line in y=mx+b form we can find the slope of the line

- myininaya

so the equation of that line can be written y=ax/4-21/4 where a/4 is the slope of this tangent line

- anonymous

The slope of the line is a right?

- anonymous

ok so then we equate the two equations together

- anonymous

right?

- myininaya

so you are talking about finding the derivative of that function y=a/x^2 at x=-2 and then setting that equal to the slope of tangent line and solving for a? that's what I'm working on right now

- anonymous

yes and I got 7 when equating the two equations together. But I don't understand how we can equate a y with a y' (derivative)

- myininaya

so we get a=a
so this equation holds for any a is what I get

- anonymous

Oh so you are making them equal to each other by solving for a?

- anonymous

a = , a =, and then equating them?

- myininaya

derivative means slope
this slope is suppose to be the same as the slope tangent to whatever point we are talkng about so we can find the slope of the tangent line and find the derivative of the curve at the given point and set them equal to find what a should be
but the resulting equation is a=a
this equation holds for every number
3=3
4=4
it is never not true

- myininaya

a/4=-2a/(x^3) at x=-2 is what I did

- anonymous

how did you get a/4?

- myininaya

##### 1 Attachment

- myininaya

a/4 is the slope of the tangent line

- anonymous

OOHHHH I understand
So we can take a/4 to be the slope, which must be the same as the graph's slope at that point?

- myininaya

right just like the very very question we did sort of

- anonymous

Got it. Thank you so much for your help. I have to go now. I really appreciate you taking your time to do this for me.

- myininaya

no problem become my fan :)

- anonymous

Sure, see you.

- myininaya

later

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