Q. 69 - Precalculus Enhanced with Graphing Utilities

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)(2,4)

Step-by-Step Solution

Answer

The values are

$$\begin{gathered} a=-\frac{4}{3} \\ b=-\frac{5}{3} \\ c=1 \end{gathered}$$

Step 1: Given information

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

Step 2: Write equations in terms of a, b,c

The equation satisfies the given points 

Hence,

$$\begin{gathered} y=a{x}^2+bx+c \\ for (x,y)=(-1,4) \\ 4=a-b+c \left(1\right) \\ For (x,y)=(2,3) \\ 3=4a+2x+c \left(2\right) \\ For (x,y)=(0,1) \\ 1=c \left(3\right) \end{gathered}$$

Step 3: Solve the equations

We get,

 From equation 3 we have $c=1$

Substitute the values in equation 1 and 2 and simplify the equations

$$\begin{gathered} a-b+1=4 \\ a-b=3 \left(4\right) \\ Similarly \\ 4a+2b+1=3 \\ 4a+2b=2 \left(5\right) \end{gathered}$$

Now multiply equation 4 by 2 and add equation 5 to it

$2a-2b=64a+2b=26a=8$

Therefore $a=\frac{4}{3}$

Now we find b

As $$\begin{gathered} a-b=3 \\ b=-3+a \\ b=-3+\frac{4}{3} \\ b=-\frac{5}{3} \end{gathered}$$

Step 4: Conclusion

The values of a, ,b, c are $\frac{4}{3},-\frac{5}{3},1$respectively.