Question

Find real numbers a, b, and c so that the

graph of the function $y=a{x}^2+bx+c$contains the points (-1, -2), (1, -4), and (2, 4).

Short answer

The values of a, ,b, c are 3, -1 ,-6

Step-by-step solution

Given information

We are given a equation$y=a{x}^2+bx+c$

Find the equations

The equation satisfies the points Therefore

$$\begin{gathered} for(x,y)=(-1,-2) \\ -2=a-b+c\left(1\right) \\ for(x,y)=(1,-4) \\ -4=a+b+c\left(2\right) \\ for(x,y)=(2,4) \\ 4=4a+2b+c\left(3\right) \end{gathered}$$

Solve the equation 

To solve the equation

Add equation 1 and 2

$a-b+c=-2+a+b+c=-42a+2c=-6$

hence we have

$$\begin{gathered} 2a+2c=-6 \\ a+c=-3 \end{gathered}$$ (4)

Now multiply the equation 1 by 2 and add it to equation 3

$2a-2b+2c=-4+4a+2b+c=46a+3c=0$

Hence $$\begin{gathered} 6a+3c=0 \\ 2a+c=0\left(5\right) \end{gathered}$$

Subtract equation 4 and 5 we get

$a+c=-3-2a-c=0-a=-3$

Therefore $a=3$

Find the values of b ,c

$$\begin{gathered} a+c=-3 \\ 3+c=-3 \\ c=-6 \end{gathered}$$

Similarly

$$\begin{gathered} a-b+c=-2 \\ 3-b-6=-2 \\ b=-1 \end{gathered}$$

Conclusion

The values of a, ,b, c are 3,-1,-6 respectively