Question

for 1 and 2, find an equation in standard form of the parabola passing through the points. to start, substitute the (x,y) values in y = ax²+bx + c to write a system of equations.

1. (2, -20), (-2, -4), (0, -8)
2. (2, 1), (1, -1), (4, -7)

Explanation:

Step1: Substitute points into equation

For the first problem, substituting the points $(2, - 20),(-2,-4),(0, - 8)$ into $y = ax^{2}+bx + c$ gives:
When $x = 2,y=-20$: $-20 = 4a+2b + c$; when $x=-2,y = - 4$: $-4=4a-2b + c$; when $x = 0,y=-8$: $-8=c$.

Step2: Solve the system of equations

Substitute $c=-8$ into the first two - equations:
$-20 = 4a+2b-8\Rightarrow4a + 2b=-12$; $-4=4a-2b-8\Rightarrow4a-2b = 4$.
Add these two new - equations: $(4a + 2b)+(4a-2b)=-12 + 4$, $8a=-8$, so $a=-1$.
Substitute $a=-1$ into $4a + 2b=-12$: $-4+2b=-12$, $2b=-8$, $b=-4$.
The equation is $y=-x^{2}-4x - 8$.

For the second problem, substituting the points $(2,1),(1,-1),(4,-7)$ into $y = ax^{2}+bx + c$ gives:
When $x = 2,y = 1$: $1 = 4a+2b + c$; when $x = 1,y=-1$: $-1=a + b + c$; when $x = 4,y=-7$: $-7=16a+4b + c$.
The system of equations is

$$\begin{cases}4a+2b + c=1\\a + b + c=-1\\16a+4b + c=-7\end{cases}$$

Answer:

1. $y=-x^{2}-4x - 8$

2.

$$\begin{cases}4a+2b + c=1\\a + b + c=-1\\16a+4b + c=-7\end{cases}$$