Question

1. 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:

The equation is $y=2x^2+4x-7$

Step-by-step explanation:

Let the quadratic function be

$y=ax^2+bx+c$

The point (-4,9) must satisfy this function,

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

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

The point (0,-7) must also satisfy this function,

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

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

The point (1,-1) must also satisfy this function,

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

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

We put equation 2 into equation 1 to get;

$\Rightarrow 16a-4b-7=9$

$\Rightarrow 16a-4b=16$

$\Rightarrow 4a-b=4...(5)$

We again put equation 2 into equation 3 to get;

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

$\Rightarrow a+b=6...(6)$

We add equation 5 and 6 to get;

$5a=10$

$\Rightarrow a=2$

We put $a=2$ into equation 6 to get;

$2+b=6$

$\Rightarrow b=6-2$

$\Rightarrow b=4$

The equation is therefore $y=2x^2+4x-7$