Question

Given three points \[\left( {0,3} \right),\left( {1, - 4} \right),\left( {2, - 9} \right)\] how do you write a quadratic function in standard form with the points?

Answer 1

Hint: The given question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the general form of a quadratic equation. Assume the given three points as \[\left( {x,y} \right)\] . We need to know how to solve the two equations with the use of arithmetic operations. First, we need to find the value of \[a,b\] and \[c\] in the quadratic equation with the use of three points given in the question.

Complete step-by-step answer:
There are three points given the question which is shown below,
 \[\left( {0,3} \right),\left( {1, - 4} \right),\left( {2, - 9} \right)\]
We have to assume the above-mentioned points as \[\left( {x,y} \right)\] . By using these points we can make the following tabular column,

x	0	1	2
y	3	-4	-9

That means,
Case \[\left( 1 \right)\] : When \[x = 0,y = 3\]
Case \[\left( 2 \right)\] : When \[x = 1,y = - 4\]
Case \[\left( 3 \right)\] : When \[x = 2,y = - 9\]
We know that,
The general form of a quadratic equation is \[a{x^2} + bx + c = y \to \left( 1 \right)\]
Let’s substitute Case \[\left( 1 \right)\] values in the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to a{x^2} + bx + c = y,\] Here, \[x = 0,y = 3\]
 \[
\Rightarrow a{\left( 0 \right)^2} + b\left( 0 \right) + c = 3 \\
  c = 3 \to \left( 2 \right) \;
 \]
So we get the value of \[c\] is \[3\] .
Let’s substitute the Case \[\left( 2 \right)\] value in the equation \[\left( 1 \right)\] we get,
 \[\left( 1 \right) \to a{x^2} + bx + c = y,\] Here, \[x = 1,y = - 4\]
 \[\Rightarrow a{\left( 1 \right)^2} + b\left( 1 \right) + c = - 4\]
 \[a + b + c = - 4\]
From the equation \[(2)\] we have \[c = 3\] . So, the above equation become
 \[\Rightarrow a + b + 3 = - 4\]
 \[a + b = - 4 - 3\]
 \[a + b = - 7 \to \left( 3 \right)\]
Let’s substitute Case \[\left( 3 \right)\] values in the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to a{x^2} + bx + c = y,\] Here, \[x = 2,y = - 9\]
 \[\Rightarrow a{\left( 2 \right)^2} + b\left( 2 \right) + c = - 9\]
 \[\Rightarrow a(4) + b(2) + c = - 9\]
 \[4a + 2b + c = - 9\]
We know that the value of \[c = 3\] .so, we get,
 \[
\Rightarrow 4a + 2b + 3 = - 9 \\
  4a + 2b = - 9 - 3 \\
  4a + 2b = - 12 \to \left( 4 \right) \\
   \;
 \]
Let’s solve the equation \[(3)\;and\;(4)\] to find the values of \[a\] and \[b\]
 \[
 \left( 3 \right) \to a + b = - 7 \\
  (4) \to 4a + 2b = - 12 \;
 \]
For easy calculation, we have to multiply the equation \[\left( 3 \right)\] with \[2\] . So, we get
 \[\left( 3 \right) \times 2 \to 2a + 2b = - 14\]
 \[(4) \to 4a + 2b = - 12\]
(Here, we use subtraction operation)
 \[ - 2a = - 2\]
 \[
\Rightarrow a = \dfrac{{ - 2}}{{ - 2}} \\
  a = 1 \;
 \]
Let’s substitute the value of \[a\] in the equation \[\left( 4 \right)\] , we get
 \[(4) \to 4a + 2b = - 12\]
 \[4\left( 1 \right) + 2b = - 12\]
 \[2b = - 12 - 4\]
 \[
\Rightarrow b = \dfrac{{ - 16}}{2} \\
  b = - 8 \;
 \]
So, let’s substitute \[a = 1,b = - 8,c = 3\] in the equation \[\left( 1 \right)\] we get,
 \[
\Rightarrow y = a{x^2} + bx + c \\
\Rightarrow y = \left( 1 \right){x^2} + \left( { - 8} \right)x + 3 \\
\Rightarrow y = {x^2} - 8x + 3 \;
 \]
So, the final answer is \[y = {x^2} - 8x + 3\]
So, the correct answer is “\[y = {x^2} - 8x + 3\] ”.

Note: When substituting a value in the quadratic equation remind the following things,
1) When a negative number is multiplied with another negative number then the answer will be positive.
2)When a positive number is multiplied with another positive number then the answer will be positive.
3)When a negative number is multiplied with the positive number then the answer will be negative.