Q. 73 - Precalculus Enhanced with Graphing Utilities

Question

Curve Fitting Find the function $y=a{x}^2+bx+c$ whose graph contains the points $\left(1,2\right),\left(-2,-7\right)$ and $\left(2,-3\right)$.

Step-by-Step Solution

Answer

The graph will be

Step 1 : Given information

The given points are $\left(1,2\right),\left(-2,-7\right)$ and $\left(2,-3\right)$.

Step 2 : Application

We require that the three points satisfy the equation $y=a{x}^2+bx+c$.

For the point $\left(1,2\right)$ :

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

$2=a+b+c$

For the point $\left(-2,-7\right)$ :

$-7=a{\left(-2\right)}^2+b\left(-2\right)+c$

$-7=4a-2b+c$

For the point $\left(2,-3\right)$:

$-3=a{\left(2\right)}^2+b\left(2\right)+c$

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

Now, determine $a$, $b$, and $c$ so that each equation is satisfied.

Solve the following system of three equations containing three variables:

$$\begin{gathered} a+b+c=2 ...\left(1\right) \\ 4a-2b+c=-7 ...\left(2\right) \\ 4a+2b+c=-3 ...\left(3\right) \end{gathered}$$

Step 3 : Solve the equations

Substracting equation (2) from equation (3),

$4b=4$

So, $b=1$

Adding equation (2) and equation (3),

$$\begin{gathered} 8a+2c=-10 \\ 4a+c=-5 ...\left(4\right) \end{gathered}$$

By equation $\left(1\right)\times 2+$equation(2),

$$\begin{gathered} 6a+3c=-3 \\ 2a+c=-1 ...\left(5\right) \end{gathered}$$

By substracting equation (5) and equation (4),

$$\begin{gathered} 2a=-4 \\ a=-2 \end{gathered}$$

Substituting value of $a$ and $b$ in the equation (1),

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

So, the quadratic function whose graph contains the points $\left(1,2\right),\left(-2,-7\right)$ and $\left(2.-3\right)$ is

$y=-2x^2+x+3$.

This is the equation of  the curve.

Step 4 : Drawing curve

The graph will be