Question

How do you find the quadratic function $y = a{x^2} + bx + c$ whose graph passes through the given points $(1,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, $( - 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} 5)$, $(2,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10)$?

Answer 1

Hint: We have to find the quadratic function using the three given points. A quadratic function is a function of degree 2. As the graph of the function passes through these points we can say that these points should also satisfy the given equation. We have to find the values of $a$, $b$ and $c$ to find the required function.

Complete step by step solution:
We are given that the graph of the quadratic function $y = a{x^2} + bx + c$ passes through the points $(1,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, $( - 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} 5)$ and $(2,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10)$.
Therefore, these points should satisfy the equation of the function. Using these three points we will find three conditions relating $a$, $b$ and $c$.
So, using point $(1,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, we can write,
$
  y = a{x^2} + bx + c \\
   \Rightarrow - 1 = a \times {1^2} + b \times 1 + c \\
   \Rightarrow - 1 = a + b + c \\
 $
Let us say this equation as condition (1)
Similarly, using second point $( - 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} 5)$, we can write,
$
  y = a{x^2} + bx + c \\
   \Rightarrow 5 = a \times {( - 1)^2} + b \times ( - 1) + c \\
   \Rightarrow 5 = a - b + c \\
 $
Let us say this equation as condition (2)
And, using third point $(2,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10)$, we can write,
$
  y = a{x^2} + bx + c \\
   \Rightarrow 10 = a \times {2^2} + b \times 2 + c \\
   \Rightarrow 10 = 4a + 2b + c \\
 $
Let us say this equation as condition (3)
We get three equations or conditions in terms of $a$, $b$ and $c$ using which we can find the value of $a$, $b$ and $c$.
From first condition we can write,
\[
   \Rightarrow - 1 = a + b + c \\
   \Rightarrow a = - 1 - b - c \\
 \]
Putting this in condition (2) we get,
$
  5 = a - b + c \\
   \Rightarrow 5 = ( - 1 - b - c) - b + c \\
   \Rightarrow 5 = - 1 - 2b \\
   \Rightarrow 2b = - 1 - 5 = - 6 \\
   \Rightarrow b = - 3 \\
 $
Putting \[a = - 1 - b - c\] and $b = - 3$ in condition (3), we get,
$
  10 = 4a + 2b + c \\
   \Rightarrow 10 = 4 \times ( - 1 - b - c) + 2b + c \\
   \Rightarrow 10 = - 4 - 4b - 4c + 2b + c \\
   \Rightarrow 10 = - 4 - 2b - 3c \\
   \Rightarrow 10 = - 4 - 2 \times ( - 3) - 3c \\
   \Rightarrow 10 = - 4 + 6 - 3c \\
   \Rightarrow 3c = - 10 - 4 + 6 = - 8 \\
   \Rightarrow c = \dfrac{{ - 8}}{3} \\
 $
Now we can find the value of $a$ as,
\[
  a = - 1 - b - c \\
   \Rightarrow a = - 1 - ( - 3) - \dfrac{{ - 8}}{3} \\
   \Rightarrow a = - 1 + 3 + \dfrac{8}{3} = \dfrac{{ - 3 + 9 + 8}}{3} \\
   \Rightarrow a = \dfrac{{14}}{3} \\
 \]
Thus, we get the values \[a = \dfrac{{14}}{3}\], $b = - 3$ and $c = - \dfrac{8}{3}$.
Using these values we can write the required quadratic function as,
$y = \dfrac{{14}}{3}{x^2} - 3x - \dfrac{8}{3}$

Hence, the quadratic function passing through the three given points is $y = \dfrac{{14}}{3}{x^2} - 3x - \dfrac{8}{3}$

Note: We used the three given points to find the three conditions relating $a$, $b$ and $c$. The graph of a quadratic function is parabolic. We can check the result by plotting the graph and locating the points or also by simply putting the values of points in the equation of the function.