A Question About a Parabola

**ty2391** · November 17th 2006, 06:40 PM

First of all, this is my first post here, and I hope to fit in with y'all here with Math.
I'm in grade 10 Math in Canada, and I just recently wrote a test on the Quadratic Function.
I didn't do as great as I wanted to, as I an usually a 90+ student. One question on the test really puzzled me, and I was shoping someone could help me with it.

Determine the equation of the parabola with x-intercepts -6 and 2, and a y intercept of 4. Express your answer in standard form.

Thanks!

**earboth** · November 17th 2006, 08:38 PM

Originally Posted by ty2391

First of all, this is my first post here, and I hope to fit in with y'all here with Math.
I'm in grade 10 Math in Canada, and I just recently wrote a test on the Quadratic Function.
I didn't do as great as I wanted to, as I an usually a 90+ student. One question on the test really puzzled me, and I was shoping someone could help me with it.

Determine the equation of the parabola with x-intercepts -6 and 2, and a y intercept of 4. Express your answer in standard form.

Thanks!

Hello, Ty2391,

the equation of your parabola is:
$y=a\cdot (x+6)(x-2)$ . Expand the RHS:

$y=ax^2+4ax-12a$ . Now you know that the y-intercept is: (0, -12a)
$-12a = 4\ \Longleftrightarrow\ a=-\frac{1}{3}$

Thus your equation is:

$y=-\frac{1}{3}x^2-\frac{4}{3}x+4$

EB

**Soroban** · November 19th 2006, 06:23 AM

Hello, ty2391!

Another approach . . .

Determine the equation of the parabola with x-intercepts -6 and 2, and a y-intercept of 4.

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

Since the y-intercept is 4, the point $(0,4)$ satisfies the equation.
. . We have: . $4 \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad c = 4$

The equation (so far) is: . $y \:=\:ax^2 + bx + 4$

Here is a sketch of the parabola . . .

Code:

                      |
                  *   |
              *   :   *
            *     :   | *
           *      :   |  *
                  :   |
      ----*-------+---+---*----
         -6      -2   |   2
                      |

We know that the axis of the parabola is halfway between the x-intecepts.
. . That is, the vertex lies on the vertical line $x = -2$

We also know that the vertex is at $x \:=\:\frac{\text{-}b}{2a}$
. . Then: . $\frac{\text{-}b}{2a}\:=\:-2\quad\Rightarrow\quad b \,= \,4a$

So the equation is: . $y \:=\:ax^2 + 4ax + 4$

Since $(2,0)$ satisfies the equation, we have:
. . $0 \:=\:a\cdot2^2 + 4a\cdot2 + 4\quad\Rightarrow\quad 12a \,=\,\text{-}4\quad\Rightarrow\quad a = \text{-}\frac{1}{3}$

Therefore, the equation is: . $\boxed{y \;= \;-\frac{1}{3}x^2 - \frac{4}{3}x + 4}$

**ty2391** · November 19th 2006, 06:52 AM

Thanks a lot both of you!

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