Question

eqn of the @ tangent lines to y=x^2 that pass thru (1,-3)?plz:)

Answer 1

2 tangent lines..sry

Answer 2

y=2x-5

Answer 3

tangent means the slope is the derivative of the function so:
derivative of x^2 = 2x... the tangent line using point slope for is then:

y-1=2(x-3)

Answer 4

*y-1=2(x+3)

I didnt see the -3

Answer 5

ohhok...wat wud the 2nd eqn b?

Answer 6

no second equation

Answer 7

i am not sure how there is a second equation.

Answer 8

but alley u switched ur y and x

Answer 9

since its a parabola..wud der b one with the opp slope r sumthing? maybe?:)

Answer 10

cuz the questions asks for 2:(

Answer 11

is there another point?

Answer 12

she only gave one

Answer 13

wow sorry got it

Answer 14

u kno how 2 get the 2nd eqn?!?!

Answer 15

no theres more to the problem than i first thought

Answer 16

so..what is it?

Answer 17

the other equation would have the opposite slope, -2, so its

y-1=-2(x+3)

Answer 18

ohh..so thts all we do?

Answer 19

alley, does that equation that you gave pass through (1,-3)?

Answer 20

sanhitha, no, it's much more complicated than that.

Answer 21

ohh really...help me out!:)

Answer 22

There should be two lines that pass through (1,-3) that are tangent to the curve. You need to figure out the general equation of the slope of a line between the point and any point on the parabola. Then you need to find the two x values where this is equal to the slope of the parabola (first derivative). Then you can determine the equations of the lines connecting the point to the 2 other points on the parabola.

Answer 23

\[y=x ^{2}\]
y'=2x

For any point (a,b) on the graph, the formula for the tangent line is therefore

(y-b)=2a(x-a) since at all x=a, the slope of the tangent line is 2a.

Since our curve is
\[y=x ^{2}\]
at the point (a,b) we have \[b=a ^{2}\]

So now, by substitution,

y-b=2a(x-a) becomes

y-a^2=2ax-2a^2

or y=2ax-a^2

So now we have a tangent line for the point (a, a^2)

So when does this line go through (1,-3)?

-3=2a(1)-a^2

a^2-2a-3=0

(a-3)(a+1)=0

a=3, -1

Plug in to our equation for the tangent line:

y=2ax-a^2

y=6x-9
y=-2x-1

And those are our answers. Notice that both pass through (1,-3) as expected.

Answer 24

ohh..finally i understood it! thank u soooo much!:)