Q. 55 - Precalculus Enhanced with Graphing Utilities

Question

Find the quadratic function $y=a{x}^2+bx+c$ that passes through the three points $(0,1),(1,0)$ and $(-2,1)$

Step-by-Step Solution

Answer

The quadratic function that passes though the points $(0,1),(1,0),(-2,1)$ is $\frac{1}{3}{x}^2+\frac{2}{3}x-1=0$

Step 1. Given

The points:

$(0,1),(1,0),(-2,1)$

A quadratic function:

$a{x}^2+bx+c=0$

Step 2. Substitute the first point in the given equation

For the point to pass through the function, the point must satisfy the function.

So for $(0,1)$,

$$\begin{gathered} y=a{x}^2+bx+c \\ 1=a{\left(0\right)}^2+b\left(0\right)+c \\ c=1 \end{gathered}$$

Step 3. Substitute the second point in the given equation

Substitute the point $(1,0)$ in the equation, we get

$$\begin{gathered} 0=a{\left(1\right)}^2+b\left(1\right)+c \\ 0=a+b+c \end{gathered}$$

Step 4. Substitute the third point in the given equation

Substitute the point $(-2,1)$ in the equation, we get

$$\begin{gathered} 1=a{(-2)}^2+b(-2)+c \\ 1=4a-2b+c \end{gathered}$$

Step 5. Write the system of equations.

The system of equations are $\left\{\begin{array}{l} c=1\\ a+b+c=0\\ 4a-2b+c=1\end{array}\right.$

Step 6. Substitute c = 1

By substituting $c=1$ in the second and third equations, we get,

$$\begin{gathered} a+b+1=0 \\ a+b=-1 \\ 4a-2b+1=1 \\ 4a-2b=0 \end{gathered}$$

Step 7 Solve the equation by substitution method.

Since 

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

Substitute $a=-1-b$ in the above equation, we get,

$$\begin{gathered} 4a-2b=0 \\ 4(-1-b)-2b=0 \\ -6b-4=0 \\ -6b=4 \\ b=-\frac{2}{3} \end{gathered}$$

The value of $b$ is $-\frac{2}{3}$

Step 8. Find a

Substitute $b=-\frac{2}{3}; c=1$ in the equation,

$$\begin{gathered} a+b+c=0 \\ a+(-\frac{2}{3})+1=0 \\ a+\frac{1}{3}=0 \\ a=-\frac{1}{3} \end{gathered}$$

The value of $a$ is $-\frac{1}{3}$

Step 9. Find the equation

Substitute the values of a, b, c in the given equation, we get,

$$\begin{gathered} a{x}^2+bx+c=0 \\ -\frac{1}{3}{x}^2+(-\frac{2}{3}x)+1=0 \\ \frac{1}{3}{x}^2+\frac{2}{3}x-1=0 \end{gathered}$$