# Math Help - parabola tangent

1. ## parabola tangent

a straight line passing through points (1,-2) is tangent of the parabola y=x^2, find the equation of the straight line.

and also what is meant by the term 'linear factors'? please explain it in general and in terms of 12x^2 + 7x - 12

and also explain how this works y = 2(2-x)^2 + 3 i got up to: -2(x-2)^2 + 3

the book however sketched the graph without it being negative.

Thanks alot

2. Originally Posted by andrew2322
and also what is meant by the term 'linear factors'? please explain it in general and in terms of 12x^2 + 7x - 12
when do you say if a polynomial is linear? so, linear factors are those factors which are linear polynomials..
for example, if you have $x^2 + 5x + 6$, then the linear factors are $x + 3$ and $x + 2$..
that is, $x^2 + 5x + 6 = (x+3)(x+2)$..

Originally Posted by andrew2322
and also explain how this works y = 2(2-x)^2 + 3 i got up to: -2(x-2)^2 + 3

the book however sketched the graph without it being negative.

Thanks alot
no.. take note that if you factor the negative there, $(2-x)^2$, you should have $(-(x-2))^2 = (-1)^2(x-2)^2$ and hence $(x-2)^2$, that is, there is no negative sign..

3. ## hey

hey kalagota, thanks for the reply however i still dont understand how it becomes one? how come u just took the negative out and put them in brackets with alone, how come it didnt end up with the 2?

consider this example 400(x-x^2)
= 400x(1-x)

in this case the x was factored out and just put infront of the 400

thanks alot.

4. Originally Posted by andrew2322
a straight line passing through points (1,-2) is tangent of the parabola y=x^2, find the equation of the straight line.
i suppose you want the non-calculus way of doing this

there will be two solutions here. see the graph below.

Let the line be $y = mx + b$

since it passes through (1,-2), we know that $-2 = m + b$ ...........(1)

Also, since the line is tangent to $y = x^2$, it means they intersect at one point. thus, there is one solution to

$x^2 = mx + b$

$\Rightarrow x^2 - mx - b = 0$

$\Rightarrow x = \frac {m \pm \sqrt{m^2 + 4b}}2$

since we want one root, we must have $m^2 + 4b = 0$ ...............(2)

Now, by (1), $b = -m - 2$, plug this into (2), we get:

$m^2 + 4(-m - 2) = 0$

$\Rightarrow m^2 - 4m - 8 = 0$

$m = \frac {4 \pm \sqrt{48}}2$

$\Rightarrow \boxed{m = 2 \pm \sqrt{12}}$

if $m = 2 + \sqrt{12}$, then $b = -4 - \sqrt{12}$

so, our line is: $\boxed{y = (2 + \sqrt{12})x + (-4 - \sqrt{12})}$

if $m = 2 - \sqrt{12}$, then $b = \sqrt{12} - 4$

so, our line is: $\boxed{y = (2 - \sqrt{12})x + (\sqrt{12} - 4)}$

taking either of these (ugly) lines work

5. Originally Posted by andrew2322
hey kalagota, thanks for the reply however i still dont understand how it becomes one? how come u just took the negative out and put them in brackets with alone, how come it didnt end up with the 2?

consider this example 400(x-x^2)
= 400x(1-x)

in this case the x was factored out and just put infront of the 400

thanks alot.
what is the square of negative numbers? in particluar, what is $(-1)^2$?

note that, $(ab)^n = a^nb^n$..

in your example, $400(x-x^2) = 400x(1-x)$ and if you take out the negative, it will become $400x(-1)(x-1) = -400x(x-1)$

considering your problem, $y = 2(2-x)^2 + 3 = 2(-(x-2))^2 + 3 = 2(-1)^2(x-2)^2 + 3$ and since the square of negative 1 is 1, therefore, you will get what the book says..