Example 9 Write a Quadratic Function Given Points Write a quadratic function for the graph that contains $$ (-7,0),(-4,-30) $$, and $$ (1,0) $$

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To write a quadratic function given three points, we can use the general form of a quadratic function:
$$ f(x) = ax^2 + bx + c $$

Given the points $$ (-7,0),(-4,-30) $$, and $$ (1,0) $$, we can substitute these points into the general form of the quadratic function to find the values of $$ a $$, $$ b $$, and $$ c $$.

Let's substitute the points into the general form of the quadratic function:
1. For the point $$ (-7,0) $$:
$$ 0 = a(-7)^2 + b(-7) + c $$
$$ 0 = 49a - 7b + c $$

2. For the point $$ (-4,-30) $$:
$$ -30 = a(-4)^2 + b(-4) + c $$
$$ -30 = 16a - 4b + c $$

3. For the point $$ (1,0) $$:
$$ 0 = a(1)^2 + b(1) + c $$
$$ 0 = a + b + c $$

Now, we have a system of three equations with three unknowns. We can solve this system to find the values of $$ a $$, $$ b $$, and $$ c $$ and then write the quadratic function.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}0=49a-7b+c\-30=16a-4b+c\0=a+b+c\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}0=49a-7b+c\-30=16a-4b+c\a=-b-c\end{array}\right.$$
- step2: Substitute the value of $$a:$$
  $$\left\{ \begin{array}{l}0=49\left(-b-c\right)-7b+c\-30=16\left(-b-c\right)-4b+c\end{array}\right.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}0=-56b-48c\-30=-20b-15c\end{array}\right.$$
- step4: Solve the equation:
  $$\left\{ \begin{array}{l}b=-\frac{6c}{7}\-30=-20b-15c\end{array}\right.$$
- step5: Substitute the value of $$b:$$
  $$-30=-20\left(-\frac{6c}{7}\right)-15c$$
- step6: Multiply the terms:
  $$-30=\frac{120c}{7}-15c$$
- step7: Swap the sides:
  $$\frac{120c}{7}-15c=-30$$
- step8: Multiply both sides of the equation by LCD:
  $$\left(\frac{120c}{7}-15c\right)\times 7=-30\times 7$$
- step9: Simplify the equation:
  $$15c=-210$$
- step10: Divide both sides:
  $$\frac{15c}{15}=\frac{-210}{15}$$
- step11: Divide the numbers:
  $$c=-14$$
- step12: Substitute the value of $$c:$$
  $$b=-\frac{6\left(-14\right)}{7}$$
- step13: Substitute back:
  $$b=\frac{6\times 14}{7}$$
- step14: Calculate:
  $$b=12$$
- step15: Substitute the values of $$b,c:$$
  $$a=-12-\left(-14\right)$$
- step16: Substitute back:
  $$a=-12+14$$
- step17: Calculate:
  $$a=2$$
- step18: Calculate:
  $$\left\{ \begin{array}{l}a=2\b=12\c=-14\end{array}\right.$$
- step19: Check the solution:
  $$\left\{ \begin{array}{l}a=2\b=12\c=-14\end{array}\right.$$
- step20: Rewrite:
  $$\left(a,b,c\right) = \left(2,12,-14\right)$$
The solution to the system of equations is $$ a = 2 $$, $$ b = 12 $$, and $$ c = -14 $$.

Therefore, the quadratic function that passes through the points $$ (-7,0),(-4,-30) $$, and $$ (1,0) $$ is:
$$ f(x) = 2x^2 + 12x - 14 $$
The quadratic function is $$ f(x) = 2x^2 + 12x - 14 $$.

Example 9 Write a Quadratic Function Given Points Write a quadratic function for the graph that contains \( (-7,0),(-4,-30) \), and \( (1,0) \)

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