Question

write a quadratic function in standard form whose graph passes through the given points1. (-1,5), (0,3), (3,9)2. (1,2), (3,4), (6,-8)

Answer 1

We have the following:

We must find the equation in each case, first calculating the slope and then the intercept with y, and finally we go to the standard form

The slope is:

$m=\frac{y_2-y_1}{x_2-x_1}$

1. (-1,5), (0,3), (3,9)

$m=\frac{9-5}{3-(-1)}=\frac{4}{3+1}=\frac{4}{4}=1$

now, for intercept with y:

x = 0, y = 3

m = 1

$\begin{gathered} y=mx+b \\ 3=1\cdot0+b \\ b=3 \end{gathered}$

The standard form:

$\begin{gathered} y=x+3 \\ y-x=3 \end{gathered}$

2. (1,2), (3,4), (6,-8)​

$m=\frac{-8-2}{6-1}=\frac{-10}{5}=-2$

now, for intercept with y:

x = 3, y =4

m = -2

$\begin{gathered} y=mx+b \\ 4=-2\cdot3+b \\ b=4+6 \\ b=10 \end{gathered}$

The standard form:

$\begin{gathered} y=-2x+10 \\ y+2x=10 \end{gathered}$