# Any Position of the Lines before Curves

##### Definition

The position of the line  $g:\quad&space;y=mx+n\quad&space;and\quad&space;parabolas\quad&space;h:\quad&space;y={&space;ax&space;}^{&space;2&space;}+bx+c$  have three possible position showed picture below.

For instance the line g : y = mx + n and parabolas h :  y = ax2 + bx + c.

When the equation of the line g is substituted over the parabola’s equation h, we get a new equation that is quadratic equation.

yh = yg

ax2 + bx + c  = mx + n

ax2 + bx  – mx+ c – n  = 0

ax2 + (b  – m)x + (c – n)  = 0………….a new equation quadratic

The Discriminant (D) of the new equation is:

D = (b – m)2 – 4a(c – n)

By seeing the value of the discriminant’s equation quadratic, it show the position of the line  g before parabolas h without to draw it graph first. This is main criteria :

1. If D > 0, then the equation quadratic has two real values, so the line  g  across parabolas h in two any point.
2. If D = 0, then the equation quadratic has two real same values, so the line g offend the parabolas h
3. If D < 0, then the equation quadratic has no real values, so the line has no intersect nor offend to the parabolic h.
##### Example 1

(Indonesian National Test)

The graph of quadratic function is defined $f(x)={&space;x&space;}^{&space;2&space;}+bx+4$  offend the line  $y=3x+4$. The value b is fulfilled by…

First step we have to make equal both of them. As we know that  $f(x)=y$ , thus we get equation

$f(x)=y\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+bx+4=3x+4\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+bx-3x=4-4\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+(b-3)x=0\\$ {quadratic equation}

The coefficient of quadratic equation is : $\\&space;a=1;\quad&space;b=(b-3);\quad&space;c=0\quad$

The condition is the equation quadratic has two real same values, so the line  offend the parabolas

Discriminant = 0

$\\&space;D=0\quad&space;\Longrightarrow&space;\quad&space;D={&space;b&space;}^{&space;2&space;}-4ac\quad&space;\{&space;Formula\}&space;\\&space;\\&space;{&space;b&space;}^{&space;2&space;}-4ac=0\\&space;\\&space;{&space;\left(&space;b-3&space;\right)&space;}^{&space;2&space;}-4.1.0=0\\&space;\\&space;{&space;\left(&space;b-3&space;\right)&space;}^{&space;2&space;}-0=0\\&space;\\&space;{&space;\left(&space;b-3&space;\right)&space;}^{&space;2&space;}=0\\&space;\\&space;b=3\\$

So the values of b needed is b=3

##### Example 2

The line  y = –2x + 3  offend a parabolas  f(x) = x2 + (m – 1)x + 7, then the values  m requirement is …

First step we have to make equals both of them. As we know that  $f(x)=y$ , thus we get equation

$\\&space;\\&space;f(x)=y\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+\left(&space;m-1&space;\right)&space;x+7=-2x+3\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+\left(&space;m-1&space;\right)&space;x+2x+7-3=0\\&space;\\&space;{&space;x&space;}^{&space;2&space;}+\left(&space;m+1&space;\right)&space;x+4=0\\&space;\\&space;\\$

The coefficient of quadratic equation is : $\\&space;a=1;\quad&space;b=m+1;\quad&space;c=4\\$

The condition is the equation quadratic has two real same values, so the line  offend the parabolas

Discriminant = 0

$D=0\\&space;\\&space;{&space;b&space;}^{&space;2&space;}-4ac=0\\&space;\\&space;{&space;\left(&space;m+1&space;\right)&space;}^{&space;2&space;}-4.1.4=0\\&space;\\&space;\left(&space;{&space;m&space;}^{&space;2&space;}+2m+1&space;\right)&space;-16=0\\&space;\\&space;{&space;m&space;}^{&space;2&space;}+2m-15=0\\$

then we get a quadratic equation in variable m. By using factor’s solving we get

$\\&space;{&space;m&space;}^{&space;2&space;}+2m-15=0\\&space;\\&space;\left(&space;m+5&space;\right)&space;\left(&space;m-3&space;\right)&space;=0\\&space;\\&space;{&space;m&space;}_{&space;1&space;}+5=0\quad&space;or\quad&space;{&space;m&space;}_{&space;2&space;}-3=0\\&space;\\&space;{&space;m&space;}_{&space;1&space;}=-5\qquad&space;\quad&space;\quad&space;\quad&space;{&space;m&space;}_{&space;2&space;}=3\\&space;\\&space;The\quad&space;values\quad&space;{&space;m&space;}_{&space;1,2&space;}\quad&space;=\{&space;-5,\quad&space;3\}&space;\\$

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