Strange Parabola

**^_^Engineer_Adam^_^** · January 9th 2007, 01:27 AM

Find the equation of the parabola
Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
could u please help me solve using this equation substituting the x and y values
$x^2 + Dx + Ey + F$

Because i always end up with this answer $x^2 - 4x + 3 = 0$ hence not a parabola

**earboth** · January 9th 2007, 03:36 AM

Originally Posted by ^_^Engineer_Adam^_^

Find the equation of the parabola
Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
could u please help me solve using this equation substituting the x and y values
$x^2 + Dx + Ey + F$

Because i always end up with this answer $x^2 - 4x + 3 = 0$ hence not a parabola

Hello,

(1. the x-value of the vertex must be the mean of -1 and 5, that means 2, because the first two points have the same y-value, they are situated symmetrical to the axis)

plug in the x- and y-values into the given equation:

$(-1,0):\ 0=1-D+F$
$(5,0):\ 0=25+5D+F$
$(1,8):\ 0=1+D+8E+F$

This is a system of simultaneous equations. I don't know which method you use. But the solution is: d = -4, E = 1, F = -5

Thus the equation of the parabola is:

$0=x^2-4x+y-5$

EB

**ticbol** · January 9th 2007, 03:39 AM

Originally Posted by ^_^Engineer_Adam^_^

Find the equation of the parabola
Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
could u please help me solve using this equation substituting the x and y values
$x^2 + Dx + Ey + F$

Because i always end up with this answer $x^2 - 4x + 3 = 0$ hence not a parabola

x^2 +Dx +Ey +F is not a parabola.
x^2 +Dx +Ey +F = 0 is. -----------(i)

At (-1,0),
(-1)^2 +D(-1) +E(0) +F = 0
1 -D +F = 0 ------------------(1)

At (5,0),
5^2 +D(5) +E(0) +F
25 +5D +F = 0 -------------(2)

(2) minus (1),
24 +6D = 0
So, D = -24/6 = -4 ----***
Substitute that into (1),
1 -(-4) +F = 0
So, F = -5 ------------***

At (1,8),
1^2 +D(1) +E(8) +F = 0
1 +D +8E +F = 0 ----------------(3)
Plug D and F in,
1 +(-4) +8E +(-5) = 0
8E -8 = 0
So, E = 8/8 = 1 -----***

Therefore, the parabola is x^2 -4x +y -5 = 0 --------answer.

**Soroban** · January 9th 2007, 07:48 AM

Hello, ^_^Engineer_Adam^_^!

Let's start from the beginning . . .

Find the equation of the parabola with a vertical axis
and passes through $(-1,0),\:(5,0),\:(1,8)$

The general equation of a "vertical" parabola is: . $y \:=\:ax^2 + bx + c$

$\begin{array}{ccc}(-1,0): \\ (5,0): \\ (1,8):\end{array}\begin{array}{ccc}0 \:=\:a(\text{-1})^2 + b(\text{-}1) + c \\ 0 \:=\:a(5^2) + b(5) + c \\ 8 \:=\:a(1^2) + b(1) + c\end{array}\begin{array}{ccc}\Rightarrow \\ \Rightarrow \\ \Rightarrow\end{array}\begin{array}{ccc}a - b + c \:=\:0 \\ 25a + 5b + c \:=\:0 \\ a + b + c \:=\:8\end{array}\begin{array}{ccc}(1)\\(2)\\(3)\e nd{array}$

Subtract (1) from (2): . $24a + 6b\:=\;0\;\;\;(4)$
Subtract (3) from (2): . $24a + 4b \:=\:\text{-}8\;\;(5)$

Subtract (5) from (4): . $2b = 8\quad\Rightarrow\quad\boxed{b = 4}$

Substitute into (4): . $24a + 6(4)\:=\:0\quad\Rightarrow\quad\boxed{a = -1}$

Substitute into (3): . $-1 + 4 + c \:=\:8\quad\Rightarrow\quad\boxed{c = 5}$

Therefore, the equation of the parabola is: . $\boxed{y \:=\:-x^2 + 4x + 5}$

Note: . $-x^2 + 4x + 5 \:=\:0$ is not a parabola.
. . . . . It is an equation . . . with two solutions: $x \:=\:-1,\,5$

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