# Thread: Equation of a parabola when two x-intercepts are given?

1. ## Equation of a parabola when two x-intercepts are given?

I am having trouble solving this problem. Any help would be appreciated!

A parabola has x-intercepts at (-2, 0) and (6, 0). Its maximum y-value is 12. Write the standard form of the equation of this parabola.

If its max y-value is 12, the y-value of the vertex is 12, right? I know I should be able to use one of the given x values to find what I need to write the equation, but I just don't know where to start with this...

2. The standard form of the parabola is: $\displaystyle f(x) = ax^2+bx+c$

Since $\displaystyle f(-2)= f(6) = 0$ we have $\displaystyle f(x) = c(x+2)(x-6) = c(x^2-4x-12)$ for some constant c.

We know $\displaystyle f(x)$ reaches maxima/minima when $\displaystyle f'(x) = c(2x-4) = 0$.

Can you figure out c with this information and write f(x) in standard form?

3. Hello, Frontier!

A parabola has x-intercepts at (-2, 0) and (6, 0).
Its maximum y-value is 12.
Write the standard form of the equation of this parabola.
Did you make a sketch?
Code:
            | (2,12)
|   *
*|   :    *
*   |   :       *
*    |   :        *
|   :
- - * - - + - + - - - - * - -
-2     |   2         6
From the symmetry, we see that the vertex is at (2, 12).

The general equation of a parabola is: .$\displaystyle f(x) \;=\;ax^2 + bc + c$

We know three points on the parabola: .$\displaystyle (-2,0),\;(2,12),\;(6,0)$
. . Use these point to create a system of equations.

$\displaystyle \begin{array}{ccccccccc}f(\text{-}2) = 0\!: & a(\text{-}2)^2 + b(\text{-}2) + c &=& 0 & \Rightarrow & 4a - 2b + c &=& 0 & [1] \\ f(2) = 12\!: & a(2^2) + b(2) + c &=& 12 & \Rightarrow & 4a + 2b + c &=& 12 & [2] \\ f(6) = 0\!: & a(6^2) + b(6) + c &=&0 & \Rightarrow & 36a + 6b + c &=& 0 & [3] \end{array}$

Subtract: [2] - [1]: .$\displaystyle 4b \:=\: 12 \quad\Rightarrow\quad \boxed{b \:=\:3}\;\;[4]$

Subtract: [3] - [2]: .$\displaystyle 32a + 4b \:=\: -1$

Substitute [4]: .$\displaystyle 32a + 4(3) \:=\:-12 \quad\Rightarrow\quad 32a \:=\:-24 \quad\Rightarrow\quad\boxed{ a \:=\:\text{-}\tfrac{3}{4}}\;\;[5]$

Substitute [4] and [5] into [1]: .$\displaystyle 4\left(\text{-}\tfrac{3}{4}\right) - 2(3) + c \:=\:0 \quad\Rightarrow\quad\boxed{ c \:=\:9}$

Therefore: .$\displaystyle f(x) \;=\;\text{-}\tfrac{3}{4}x^2 + 3x + 9$

Edit: Dinkydoe has a much better solution!'.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/pre-calculus/124784-equation-parabola-when-two-x-intercepts-given.html", "fetch_time": 1527448643000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00480.warc.gz", "warc_record_offset": 189476721, "warc_record_length": 10418, "token_count": 900, "char_count": 2403, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "latest", "language": "en", "language_score": 0.7503979802131653, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"}