Chapter 11.1, Problem 69AYU

To determine

To find: Find real numbers $a, b,$ and $c$ so that the graph of the function $y=a{x}^2+bx+c$ contains the points (-1,4)(2,3) and (0,1)

Answer to Problem 69AYU

Solution:

  $a=\frac{4}{3},b=\frac{-5}{3},c=1$

  $y=\frac{4}{3}{x}^2\frac{-5}{3}x+1$

Explanation of Solution

Given:

The graph of the function $y=a{x}^2+bx+c$ contains the points (-1.4),(2,3) and (0,1)

Calculation:

The three points must satisfy the equation $y=a{x}^2+bx+c$

For the point (-1,4) we have,

  $y=a{x}^2+bx+c$

  $4=a{(-1)}^2+b(-1)+c$

  $4=a-b+c-----1$

For the point (2,3) we have,

  $y=a{x}^2+bx+c$

  $3=a{(2)}^2+b(2)+c$

  $3=4a+2b+c----2$

For the point (0,1) we have,

  $y=a{x}^2+bx+c$

  $1=a{(0)}^2+b(0)+c$

  $c=1$

Substituting $\text{c}=1$ we have,

  $4=a-b+1$

  $a-b=3----3$

  $3=4a+2b+1$

  $4a+2b=2$

  $2a+b=1---4$

Adding equation 3 and 4 we have,

  $3a=4$

  $a=\frac{4}{3}$

Substituting value of a in equation 3 we have,

  $a-b=3$

  $\frac{4}{3}-b=3$

  $b=\frac{-5}{3}$

Therefore, $a=\frac{4}{3}xb=\frac{-5}{3},c=1$ .