two tangents to this graph....find the euation of both

[wind]

Sketch the graph of y=x^2 and show the two tangents to this graph that pass throught the point ( 0 , -4). Find the euation of both tangents.

y - y1 = m(x - x1)
y + 4 = m(x - 0)
y = mx - 4


But how do I find the slope?

 

[wjm11]

Sketch the graph of y=x^2 and show the two tangents to this graph that pass throught the point ( 0 , -4). Find the euation of both tangents.

y - y1 = m(x - x1)
y + 4 = m(x - 0)
y = mx - 4

The derivative of the quadratic y=x^2 is f’ = 2x, so the function f(x) = 2x gives the slope of the parabola for any value of x.

Rearrange your eqn, y = mx – 4, to solve for m, the slope. Do you see that m must also equal 2x? Can you take it from here?

 

[wind]

Do you see that m must also equal 2x?

Yes, I understand that but I don't really get the rest

Rearrange your eqn, y = mx – 4, to solve for m, the slope.

Like this?

y = mx – 4
y + 4 / x = m

m must also equal 2x

y + 4 / x = 2x

.... now what can you further explain this? Thanks.

 

[galactus]

The two lines are essentially the same except they their slopes have opposite signs.

\(\displaystyle y-y_{1}=m(x-x_{1})\)

\(\displaystyle x^{2}+4=2x(x-0)\)

Solve for x. One will be negative, one positive.

Then all you have to do is enter those x values into the slope m=2x.

There you have it, y=mx-4

You know that b equals -4 for both. That's a given. That's where they intercept the y axis.

 

[wind]

Thanks