Question

Write a quadratic function, in standard form, that fits the set of points. Solve it as a system of three equations. (-4, 9), (0, -7), and (1, -1)

Answer 1

Answer:

$\boxed{y=2x^2+4x-7}$

Step-by-step explanation:

Let the quadratic function be

$y=ax^2+bx+c$

We substitute $(-4,9)$ into the equation to obtain;

$9=a(-4)^2+b(-4)+c$

$\Rightarrow 9=16a-4b+c---(1)$

We substitute $(0,-7)$ to obtain;

$-7=a(0)^2+b(0)^2+c$

$\Rightarrow c=-7---(2)$

We finally substitute $(1,-1)$ to obtain;

$-1=a(1)^2+b(1)^2+c$

$\Rightarrow -1=a+b+c---(3)$

We put equation (2) into equation (1) to get;

$9=16a-4b-7$

$16a-4b=16$

$\Rightarrow 4a-b=4---(4)$

$\Rightarrow -1=a+b-7$

$\Rightarrow a+b=6---(5)$

We add equation (4) and (5) to get;

$4a+a=6+4$

$\Rightarrow 5a=10$

$\Rightarrow a=2$

We put $a=2$ into equation (5) to get;

$2+b=6$

$\Rightarrow b=6-2$

$\Rightarrow b=4$

The reqiured quadratic function is

$y=2x^2+4x-7$