# How do you write a quadratic function in standard form whose graph passes through points (1,-1/2), (0,-3), (2,3)?

Mar 17, 2017

$f \left(x\right) = \frac{1}{2} {x}^{2} + 2 x - 3$

#### Explanation:

Notice that the distinct $x$ coordinates are $\left\{0 , 1 , 2\right\}$, which are in arithmetic progression with common difference $1$.

Let us write down the corresponding $y$ coordinates in ascending order of $x$ coordinate:

$\textcolor{b l u e}{- 3}$, -1/2, 3

Next, write down the sequence of differences between consecutive terms:

$\textcolor{b l u e}{\frac{5}{2}} , \frac{7}{2}$

Next, write down the sequence of differences between consecutive terms of that sequence:

$\textcolor{b l u e}{1}$

Having arrived at a constant sequence (albeit of just one term), we can use the first term of each of these sequences to write a formula for the desired function:

f(x) = color(blue)(-3)/(0!) + color(blue)(5/2)/(1!)x + color(blue)(1)/(2!)x(x-1)#

$\textcolor{w h i t e}{f \left(x\right)} = - 3 + \frac{5}{2} x + \frac{1}{2} {x}^{2} - \frac{1}{2} x$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{2} {x}^{2} + 2 x - 3$

graph{(y - (1/2x^2+2x-3))(2x^2+(y+3)^2-0.02)(2(x-1)^2+(y+1/2)^2-0.02)(2(x-2)^2+(y-3)^2-0.02) = 0 [-9.59, 6, -6, 4.84]}