# How do you write an equation of the quadratic function whose graph contains the points (-4,-57), (2, 39), (5, 168)?

Oct 2, 2015

$3 {x}^{2} + 6 x + 3$

#### Explanation:

General form of a quadratic equation is
$\textcolor{w h i t e}{\text{XX}} y = a {x}^{2} + b x + c$

So for the given points, we have
[1]$\textcolor{w h i t e}{\text{XX}} - 57 = a {\left(- 4\right)}^{2} + b \left(- 4\right) + c$
[2]$\textcolor{w h i t e}{\text{XX}} 39 = a {\left(2\right)}^{2} + b \left(2\right) + c$
[3]$\textcolor{w h i t e}{\text{XX}} 168 = a {\left(5\right)}^{2} + b \left(5\right) + c$

Simplifying these:
[3]$\textcolor{w h i t e}{\text{XX}} 16 a - 4 b + c = - 57$
[4]$\textcolor{w h i t e}{\text{XX}} 4 a + 2 b + c = 39$
[5]$\textcolor{w h i t e}{\text{XX}} 25 a + 5 b + c = 168$

Subtracting [4] from [3]
[6]$\textcolor{w h i t e}{\text{XX}} 12 a - 6 b = - 96$
Subtracting [4] from [5]
[7]$\textcolor{w h i t e}{\text{XX}} 21 a + 3 b = 129$

Multiplying [7] by 2
[8]$\textcolor{w h i t e}{\text{XX}} 42 a + 6 b = 258$

[9]$\textcolor{w h i t e}{\text{XX}} 54 a = 162$

Dividing [9] by 3
[10]$\textcolor{w h i t e}{\text{XX}} a = 3$

Substituting $3$ for $a$ in [6]
[11]$\textcolor{w h i t e}{\text{XX}} 12 \left(3\right) - 6 b = - 96$

Simplifying
[12]$\textcolor{w h i t e}{\text{XX}} - 6 b = - 132$

Dividing [12] by $- 6$
[13]$\textcolor{w h i t e}{\text{XX}} b = 12$

Substituting $12$ for $b$ (from [13]) and $3$ for $a$ (from [10]) in [4]
[14]$\textcolor{w h i t e}{\text{XX}} 4 \left(3\right) + 2 \left(12\right) + c = 39$

Simplifying
[15]$\textcolor{w h i t e}{\text{XX}} 12 + 24 + c = 39$

[16]$\textcolor{w h i t e}{\text{XX}} c = 3$

Oct 2, 2015

$y = 3 {x}^{2} + 22 x - 17$

#### Explanation:

First, we will create three equations, one for each point. It will be in the general form $y = a {x}^{2} + b x + c$.

First Equation from P(-4,-57)
$y = a {x}^{2} + b x + c$
$\left(- 57\right) = a {\left(- 4\right)}^{2} + b \left(- 4\right) + c$
$- 57 = 16 a - 4 b + c$

Second Equation from P(2,39)
$y = a {x}^{2} + b x + c$
$\left(39\right) = a {\left(2\right)}^{2} + b \left(2\right) + c$
$39 = 4 a + 2 b + c$

Third Equation from P(5,168)
$y = a {x}^{2} + b x + c$
$\left(168\right) = a {\left(5\right)}^{2} + b \left(5\right) + c$
$168 = 25 a + 5 b + c$

System of Equations:
$- 57 = 16 a - 4 b + c$
$39 = 4 a + 2 b + c$
$168 = 25 a + 5 b + c$

Now, we can solve for the values of $a$, $b$, and $c$. Let's try getting two more equations to help us solve this system.

Fourth Equation taken from the first and second equations:
We will first get the value of $c$ in the first equation then we will substitute it into $c$ in the second equation.
$- 57 = 16 a - 4 b + c$
$- 57 - 16 a + 4 b = 16 a - 4 b + c - 16 a + 4 b$
$- 57 - 16 a + 4 b = c$

$39 = 4 a + 2 b + c$
$39 = 4 a + 2 b + \left(- 57 - 16 a + 4 b\right)$
$39 = 4 a + 2 b - 57 - 16 a + 4 b$
$39 + 57 = - 12 a + 6 b$
$96 = - 12 a + 6 b$

Let's simplify this to:
$16 = - 2 a + b$

Fifth Equation taken from the first and third equations:
We will substitute the value of $c$ in the first equation into $c$ in the third equation.
$168 = 25 a + 5 b + c$
$168 = 25 a + 5 b + \left(- 57 - 16 a + 4 b\right)$
$168 = 25 a + 5 b - 57 - 16 a + 4 b$
$168 + 57 = 9 a + 9 b$
$225 = 9 a + 9 b$

Let's simplify this to:
$25 = a + b$

Solution
Now, we will use the 4th and 5th equation to solve for $a$ and $b$. I'll use the elimination method since it is easier:

$- 2 a + b = 16$
$\underline{- a - b = - 25}$
$- 3 a = - 9$
$a = 3$

$25 = a + b$
$25 = 3 + b$
$b = 22$

To solve for $c$ just substitute the values of $a$ and $b$ into any of the first three equations.
$39 = 4 a + 2 b + c$
$39 = 4 \left(3\right) + 2 \left(22\right) + c$
$39 = 12 + 44 + c$
$c = - 17$

So the values of $a$, $b$, and $c$ are:
$a = 3$
$b = 22$
$c = - 17$

Finally, to get the quadratic equation, just substitute the values of $a$, $b$, and $c$ into the general form $y = a {x}^{2} + b x + c$.
$y = a {x}^{2} + b x + c$
$\textcolor{red}{y = 3 {x}^{2} + 22 x - 17}$