# Math Help - I need your help bruddas! ASAP! STAT! NOW!

1. ## I need your help bruddas! ASAP! STAT! NOW!

For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3

Find the Y coordinate of the turning point.
The equation of the parabola
The coordinates of the x-intercepts.

-----------------------------------------------------------------

The parabola has a turning point at (Z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

Find the value of Z.
Find the equation of the parabola.

-----------------------------------------------------------------

The graph of $y = a(h - x)^3 + k$ cuts the x-axis at x = - 4 and the y axis and y= 28. It has a dilation of 1.

Find the position of the SPOI.
-----------------------------------------------------------------

Thanks
Help would be appreciate.

Cheers

2. Originally Posted by mibamars
For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3
What have you done?

The Range restriction is in 'y', so this must open up or down. None of that left or right stuff.

You then have y - k = a(x-h)^2

You are given k = 3 and a < 0 and h = -4

y - 3 = a(x+4)^2

Substitute x = 0 and find the value for 'a' that gives y = -7/3.

Let's see what you get.

3. Originally Posted by mibamars
For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3

Find the Y coordinate of the turning point.
The equation of the parabola
The coordinates of the x-intercepts.
Try to post one question at a time. It leads us to believe you just want questions answered for homework when you post so many at once within one thread.

I assume turning point is where the graph turns, hence where a max or min is at. This is also known as a critical point. The graph is a parabola, and it has a maximum value at $y=3$ thanks to the given range. Therefore, that maximum must be the critical point at $x=-4$. The coordinates of the critical point are then (-4, 3).

The parabola takes on this equation: $y=a(x+c)^2+b$

4. Hello, mibamars!

Here's the first one . . .

For the parabola whose range is $y \leq 3$,
whose x-coordinate of the turning point is -4,
and whose y-intercept is $-\frac{7}{3}$

Find:
(1) the y-coordinate of the turning point.
(2) the equation of the parabola
(3) the coordinates of the x-intercepts.

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

Make a sketch . . .
Code:
                      |
(-4,3)     |
*       |
*     *    |
*         *  |
*           * |
- - - - - - - - + - - - - -
*             *|
|
|
*               *
|
Since $\left(0,\,-\frac{7}{3}\right)$ is on the parabola: . $-\frac{7}{3} \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad \boxed{c \:=\:-\frac{7}{3}}$

The equation (so far) is: . $y \;=\;ax^2 + bx - \frac{7}{3}$

Since the range (values of y) is < 3,
. . and the x-coordinate of the turning point is -4,
. . clearly the vertex is (-4,3).

${\color{red}\text{(1) The }y\text{-coordinate of the turning point is }3.}$

The x-coordinate of the vertex is given by: . $\frac{-b}{2a}$
. . So we have: . $\frac{-b}{2a} \:=\:-4\quad\Rightarrow\quad 8a - b \:=\:0$ . [1]

Since (-4,3) is on the parabola: . $3 \:=\:a(\text{-}4)^2 + b(\text{-}4) - \frac{7}{3}$
. . which simplifies to: . $16a - 4b \:=\:\frac{16}{3}\quad\Rightarrow\quad 4a - b \:=\:\frac{4}{3}$ . [2]

Subtract [2] from [1]: . $4a \:=\:-\frac{4}{3}\quad\Rightarrow\quad\boxed{ a \:=\:-\frac{1}{3}}$

Substitute into [1]: . $8\left(\text{-}\frac{1}{3}\right) - b \:=\:0\quad\Rightarrow\quad\boxed{ b \:=\:-\frac{8}{3}}$

${\color{red}\text{(2) The equation of the parabola is: }\:y \;=\;-\frac{1}{3}x^2 - \frac{8}{3}x - \frac{7}{3}}$

For the x-intercepts: . $-\frac{1}{3}x^2 - \frac{8}{3}x - \frac{7}{3} \;=\;0$

Multiply by -3: . $x^2 + 8x + 7 \:=\:0$

. . Factor:. $(x+7)(x+1) \:=\:0\quad\Rightarrow\quad x \:=\:-7,-1$

${\color{red}\text{(3) The }x\text{-intercepts are: }\:(-7,0), (-1,0)}$