# How do you write a quadratic function in standard form whose graph passes through points (-1,-2), (1,-4), (2,1)?

Mar 28, 2017

$\textcolor{g r e e n}{y = 2 {x}^{2} - x - 5}$

#### Explanation:

$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$ for some constants $a , b , c$

{: ("Given point:",color(white)("XXX"),"Implied equation:"), ((-1,-2),rArr,-2=a-b+c color(white)("XX")[1]), ((1,-4),rArr,-4=a+b+c color(white)("XX")[2]), ((2,1),rArr,1=4a+2b+c color(white)("XX")[3]) :}

Subtracting [2] - [1]
$\textcolor{w h i t e}{\text{XXX")-2=2bcolor(white)(*"XX}} \left[4\right]$
$\textcolor{w h i t e}{\text{XXX")rarrcolor(white)("XX}} b = - 1$

Substituting $\left(- 1\right)$ for $b$ in [2] and [3]
$\textcolor{w h i t e}{\text{XXX")-3=a+c color(white)("XX}} \left[5\right]$
$\textcolor{w h i t e}{\text{XXXx")3=4a+c color(white)("XX}} \left[6\right]$

Subtracting [6]-[5]
$\textcolor{w h i t e}{\text{XXX}} 6 = 3 a$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow a = 2$

Substituting $2$ for $a$ in [5]
$\textcolor{w h i t e}{\text{XXX}} - 3 = 2 + c$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow c = - 5$

and the general quadratic standard equation: $y = a {x}^{2} + b x + c$
becomes
$\textcolor{w h i t e}{\text{XXX}} y = 2 {x}^{2} - 1 x - 5$

As a verification, here is the graph of $y = 2 {x}^{2} - x - 5$ and the given points: