Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

Curve Fitting Find the function $y=a{x}^2+bx+c$ whose graph contains the points $\left(1,-1\right),\left(3,-1\right)$ and $\left(-2,14\right)$.

Step-by-Step Solution

Answer

The graph is

Step 1 : Given information

The given points are $\left(1,-1\right),\left(3,-1\right)$ and $\left(-2,14\right)$.

Step 2 : Application

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

For the point $\left(1,-1\right)$ :

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

For the point  $\left(3,-1\right)$:

$$\begin{gathered} -1=a{\left(3\right)}^2+b\left(3\right)+c \\ -1=9a+3b+c \end{gathered}$$

For the point $\left(-2,14\right)$:

$$\begin{gathered} 14=a{\left(-2\right)}^2+b\left(-2\right)+c \\ 14=4a-2b+c \end{gathered}$$

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

Solve the following system of three equations containing three variables:

$$\begin{gathered} a+b+c=-1 ...\left(1\right) \\ 9a+3b+c=-1 ...\left(2\right) \\ 4a-2b+c=14 ...\left(3\right) \end{gathered}$$

Step 3 : Solve the equations

From equation (1) and (2),

$(-)a+b+c=-19a+3b+c=-1-8a-2b=0$

$$\begin{gathered} -8a=2b \\ b=-4a ...\left(4\right) \end{gathered}$$

From equation (2) and equation (3),

$(-)9a+3b+c=-14a-2b+c=145a+5b=-15$

$a+b=-3 ...\left(5\right)$

Substituting value of $b$ in the equation (5),

$$\begin{gathered} a-4a=-3 \\ -3a=-3 \\ a=1 \end{gathered}$$

From equation (4),

  $$\begin{gathered} b=-4\left(1\right) \\ b=-4 \end{gathered}$$

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

$$\begin{gathered} 1-4+c=-1 \\ c=2 \end{gathered}$$

So, the quadratic function whose graph contains the points $\left(1,-1\right),\left(3,-1\right)$ and $\left(-2,14\right)$ is

$y=x^2-4x+2$

This is the equation of the curve.

Step 4 : Drawing curve

The graph will be