Question

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

Answer 1

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
....

Answer 2

thinking...

Answer 3

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

Answer 4

right

Answer 5

Would those two lines be orthogonal to each other?

Answer 6

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

Answer 7

Ok so then they must have reciprocal slopes

Answer 8

opposite reciprocal slopes

Answer 9

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

Answer 10

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)

Answer 11

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)

Answer 12

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!

Answer 13

so the lines arent orthogonal

Answer 14

do you have any questions?

Answer 15

One question

Answer 16

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?

Answer 17

(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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

cool?

Answer 22

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

Answer 23

go for it

Answer 24

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

Answer 25

i got it scanning it now

Answer 26

i just used the definition of derivative

Answer 27

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

Answer 28

do you understand both problems we did?

Answer 29

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

Answer 30

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

Answer 31

what is definition of the derivative at a of f

Answer 32

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

Answer 33

Sorry I don't know what happened

Answer 34

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

Answer 35

oops the bottom is suppose to read 2h

Answer 36

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

Answer 37

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

Answer 38

yes

Answer 39

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?

Answer 40

yes

Answer 41

ok 6*(1/2)=3

Answer 42

Got it. Thank you.

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

The slope of the line is a right?

Answer 47

ok so then we equate the two equations together

Answer 48

right?

Answer 49

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

Answer 50

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)

Answer 51

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

Answer 52

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

Answer 53

a = , a =, and then equating them?

Answer 54

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

Answer 55

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

Answer 56

how did you get a/4?

Answer 57

a/4 is the slope of the tangent line

Answer 58

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?

Answer 59

right just like the very very question we did sort of

Answer 60

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.

Answer 61

no problem become my fan :)

Answer 62

Sure, see you.

Answer 63

later