Q. 116 - Precalculus Enhanced with Graphing Utilities

Question

Make up an equation satisfied by the ordered pairs $\left(2,0\right),\left(4,0\right),\left(0,1\right)$ Compare your equation with a friend’s equation. Comment on any similarities.

Step-by-Step Solution

Answer

The equation so formed is $y=\frac{1}{8}{x}^2-\frac{3}{4}x+1$

Step 1. Given Information

Given ordered pairs are $\left(2,0\right),\left(4,0\right) and \left(0,1\right)$.

In this exercise, lets determine the function that passes through the given points. As can be seen, the three points are not collinear because two of the points lie on the $x$-axis while the other one lies in the first quadrant. 

Step 2. Forming possible equation

Lets assume that the function is parabola.$y=a{x}^2+bx+c$.

Now, write three satisfied equation by using given coordinate at $\left(2,0\right)$ and $\left(4,0\right)$ respectively

$$\begin{gathered} \left(i\right)0=a{\left(2\right)}^2+2b+c=0\Rightarrow 4a+2b+c=0 \\ \left(ii\right)0=a{\left(4\right)}^2+4b+c=0\Rightarrow 16a+4b+c=0 \end{gathered}$$

$\left(iii\right)1=a{\left(0\right)}^2+0b+c=0\Rightarrow c=-1$

Step 3. Form linear equation by putting c = 1

$$\begin{gathered} 4a+2b=-1 \\ 16a+4b=-1 \end{gathered}$$

Step 4. Finding the values of a and b

$$\begin{gathered} 4a+2b=-1----\left(i\right) \\ 16a+4b=-1----\left(ii\right) \end{gathered}$$

Solving these two equations by elimination method, we get

$a=\frac{1}{8},b=-\frac{3}{4}$

Step 5. Put the values of a , b , c in the parabola

$$\begin{gathered} y=a{x}^2+bx+c \\ y=\frac{1}{8}{x}^2-\frac{3}{4}x+1 \end{gathered}$$