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

2 tangent lines..sry

y=2x-5

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)

*y-1=2(x+3) I didnt see the -3

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

no second equation

i am not sure how there is a second equation.

but alley u switched ur y and x

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

cuz the questions asks for 2:(

is there another point?

she only gave one

wow sorry got it

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

no theres more to the problem than i first thought

so..what is it?

the other equation would have the opposite slope, -2, so its y-1=-2(x+3)

ohh..so thts all we do?

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

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

ohh really...help me out!:)

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.

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

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

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