anonymous
  • anonymous
Determine the equations of both lines that are tangent to the graph of f(x) = x^2 and pass through the point (1,-3).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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myininaya
  • 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
  • myininaya
thinking...
anonymous
  • anonymous
One would have a negative slope, the other would be positive

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myininaya
  • myininaya
right
anonymous
  • anonymous
Would those two lines be orthogonal to each other?
myininaya
  • myininaya
that looks possible it looks like it could form a 90 degree angles
anonymous
  • anonymous
Ok so then they must have reciprocal slopes
myininaya
  • myininaya
opposite reciprocal slopes
myininaya
  • myininaya
wait i htink i'm fixing to figure something out....
myininaya
  • myininaya
1 Attachment
myininaya
  • 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
  • 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
  • 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
  • myininaya
so the lines arent orthogonal
myininaya
  • myininaya
do you have any questions?
anonymous
  • anonymous
One question
anonymous
  • 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
  • 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
  • myininaya
(y1-y2)/(x1-x2)=(y2-y1)/(x2-x1) (-1)/(-1)*(y1-y2)/(x1-x2)=(y2-y1)/(x2-x1)
myininaya
  • myininaya
a-b is the same as -(b-a)
myininaya
  • myininaya
(6-2)/(3-1)=4/2=2 (2-6)/(1-3)=-4/(-2)=2
myininaya
  • myininaya
cool?
anonymous
  • anonymous
Yes thank you very much. I have just a few more questions if you don't mind helping me out with them.
myininaya
  • myininaya
go for it
anonymous
  • anonymous
If f(a) = 0 and f'(a) = 6, find the limit h --> 0 of: f(a + h) ------- 2h
myininaya
  • myininaya
i got it scanning it now
myininaya
  • myininaya
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myininaya
  • myininaya
i just used the definition of derivative
myininaya
  • myininaya
and plugged and what they gave me and solved for limh->0 (f(a+h)/(2a))
myininaya
  • myininaya
do you understand both problems we did?
myininaya
  • myininaya
oops solved for limh->0(f(a+h)/(2h))
anonymous
  • anonymous
I don't understand how you did the second question.
myininaya
  • myininaya
what is definition of the derivative at a of f
anonymous
  • 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
  • anonymous
Sorry I don't know what happened
myininaya
  • myininaya
doesn't the question above ask for lim h->0 (f(a+h)/(2a))
myininaya
  • myininaya
oops the bottom is suppose to read 2h
myininaya
  • myininaya
so we have 6=limh->0(f(a+h)/(2h))
myininaya
  • myininaya
oops 6=limh->0(f(a+h)/h) do you understand this far?
anonymous
  • anonymous
yes
myininaya
  • 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
  • anonymous
yes
myininaya
  • myininaya
ok 6*(1/2)=3
anonymous
  • anonymous
Got it. Thank you.
anonymous
  • 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
  • myininaya
if you write that tangent line in y=mx+b form we can find the slope of the line
myininaya
  • 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
  • anonymous
The slope of the line is a right?
anonymous
  • anonymous
ok so then we equate the two equations together
anonymous
  • anonymous
right?
myininaya
  • 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
  • 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
  • myininaya
so we get a=a so this equation holds for any a is what I get
anonymous
  • anonymous
Oh so you are making them equal to each other by solving for a?
anonymous
  • anonymous
a = , a =, and then equating them?
myininaya
  • 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
  • myininaya
a/4=-2a/(x^3) at x=-2 is what I did
anonymous
  • anonymous
how did you get a/4?
myininaya
  • myininaya
1 Attachment
myininaya
  • myininaya
a/4 is the slope of the tangent line
anonymous
  • 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
  • myininaya
right just like the very very question we did sort of
anonymous
  • 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
  • myininaya
no problem become my fan :)
anonymous
  • anonymous
Sure, see you.
myininaya
  • myininaya
later

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