The parabola has the turning point at (z, -8) and y-intercept at 10, and an x-intercept at x=5.

What is --- value of z?

The equation of parabola?

Plz help!

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- Mar 24th 2008, 03:16 AMmibamarsParabola Question! Help!
The parabola has the turning point at (z, -8) and y-intercept at 10, and an x-intercept at x=5.

What is --- value of z?

The equation of parabola?

Plz help! - Mar 24th 2008, 03:48 AMmr fantastic
There are several approaches that can be taken. This is one of them:

The equation can be written in turning point form as $\displaystyle y = a(x - z)^2 - 8$.

Substitute the known points (0, 10) and (5, 0) to get two equations in the unknowns a and z. Solve these two equations simultaneously to get the values of a and z. - Mar 24th 2008, 04:04 AMmibamars
Still cant solve it. Stuck on the unknown values. Can someone do a full solve?

- Mar 24th 2008, 04:14 AMtopsquark
$\displaystyle y = a(x - z)^2 - 8$

Plug in the two points:

$\displaystyle 10 = a(-z)^2 - 8$

and

$\displaystyle 0 = a(5 - z)^2 - 8$

respectively.

The first equation says:

$\displaystyle 10 = az^2 - 8$

so

$\displaystyle a = \frac{18}{z^2}$

Inserting this into the second equation gives:

$\displaystyle 0 = \left ( \frac{18}{z^2} \right )(5 - z)^2 - 8$

$\displaystyle 0 = (5 - z)^2 - 8$

Solve this for z, then use that z value to find a.

-Dan - Mar 24th 2008, 05:23 AMmr fantastic
- Mar 24th 2008, 05:31 AMSoroban
Hello, mibamars!

Another approach . . .

Quote:

The parabola has the turning point at $\displaystyle (z,\,\text{-}8)$,

and y-intercept at $\displaystyle 10$, and an x-intercept at $\displaystyle 5.$

Find $\displaystyle z$ and the equation of the parabola.

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

We are given three points on the parabola.

$\displaystyle (0,10)\!:\;\;10 \:=\:a(0^2) + b(0) + c\quad\Rightarrow\quad\boxed{ c \:=\:10}$

$\displaystyle (5,0)\!:\;\;0 \:=\:25a + 5b + 10\quad\Rightarrow\quad 5a + b \:=\:\text{-}2\;\;{\color{blue}[1]}$

$\displaystyle (z,\text{-}8)\!:\;\;\text{-}8 \:=\:az^2 + bz + 10\quad\Rightarrow\quad az^2 + bz \:=\:\text{-}18\;\;{\color{blue}[2]}$

We are told that: .$\displaystyle y'(z) = 0$

We have: .$\displaystyle y' \:=\:2ax + b\quad\hdots\quad\text{Hence: }\; 2az + b\:=\:0\;\;{\color{blue}[3]}$

$\displaystyle \begin{array}{cccc}\text{Multiply {\color{blue}[3]} by }z\!:& 2az^2 + bz &=&0 \\

\text{Subtract {\color{blue}[2]}}\!: & az^2 + bz &=&\text{-}18 \end{array}$

And we have: .$\displaystyle az^2 \:=\:18\quad\Rightarrow\quad a \:=\:\frac{18}{z^2}\;\;{\color{blue}[4]}$

Substitute into [3]: .$\displaystyle 2\left(\frac{18}{z^2}\right)z +b \:=\:0\quad\Rightarrow\quad b \:=\:-\frac{36}{z}\;\;{\color{blue}[5]}$

Substitute [4] and [5] into [1]: .$\displaystyle 5\left(\frac{18}{z^2}\right) - \frac{36}{z}\:=\:-2\quad\Rightarrow\quad z^2 - 18z + 45\:=\:0$

. . and we have: .$\displaystyle (z-3)(z-15)\:=\:0\quad\Rightarrow\quad\boxed{z \:=\:3,\:15}$

There are__two__solutions . . .

. . $\displaystyle \begin{array}{ccccccccc}z \,=\,3, & a \,=\,2, & b\,=\,\text{-}12 && \Longrightarrow & &\boxed{ y \;=\;2x^2 - 12x + 10} \\ \\

z\,=\,15 & a \,=\,\frac{2}{25}, & b\,=\,\text{-}\frac{12}{5} && \Longrightarrow &&\boxed{ y \;=\;\frac{2}{25}x^2 - \frac{12}{5}x + 10} \end{array}$

- Mar 24th 2008, 01:40 PMtopsquark