Question

If \[f\left( x \right)\] is a polynomial function of the second degree such that, \[f\left( { - 3} \right) = 6,f\left( 0 \right) = 6\]and \[f\left( 2 \right) = 11\], then the graph of the function, \[f\left( x \right)\]cuts the ordinate \[x = 1\]at the point
A. \[\left( {1,8} \right)\]
B. \[\left( {1,4} \right)\]
C. \[\left( {1, - 2} \right)\]
D. None of these

Answer 1

Hint: In this question, we need to find the point at which the graph of the function \[f\left( x \right)\] cuts the ordinate \[x = 1\]. For this, we have to consider the polynomial function \[f\left( x \right) = a{x^2} + bx + c\]. Also, we have to use the given conditions to find the values of \[a,b\]and\[c\]. After that we will find the value of \[f\left( x \right)\].

Complete step-by-step answer:
Let us consider the polynomial function \[f\left( x \right)\] be \[a{x^2} + bx + c\].
That is \[f\left( x \right) = a{x^2} + bx + c\]
But \[f\left( 0 \right) = 6\]
So, put \[x = 0\] in the function \[f\left( x \right)\].
Thus, we get
\[f\left( 0 \right) = a{\left( 0 \right)^2} + b\left( 0 \right) + c\]
\[6 = 0 + 0 + c\]
\[6 = c\]
Now, consider \[f\left( { - 3} \right) = 6\]
So, put \[x = - 3\]in the function \[f\left( x \right)\].
Thus, we get
\[f\left( { - 3} \right) = a{\left( { - 3} \right)^2} + b\left( { - 3} \right) + c\]
\[6 = 9a - 3b + 6\]
\[6 - 6 = 9a - 3b\]
By simplifying, we get
\[0 = 9a - 3b\]
Divide by 3 to both sides, we get
\[0 = 3a - b\]…… (1)
Also, we know that \[f\left( 2 \right) = 11\]
So, put \[x = 2\] in the function \[f\left( x \right)\].
Thus, we get
\[f\left( 2 \right) = a{\left( x \right)^2} + b\left( 2 \right) + 6\]
\[11 - 6 = a{\left( 2 \right)^2} + b\left( 2 \right)\]
\[5 = 4a + 2b\]…… (2)
Let us solve equations (1) and (2) to find the values of \[a\] and \[b\]
From equation (1), we get
\[b = 3a\]
Put \[b = 3a\]in the equation (2)
So, we get
\[5 = 4a + 2\left( {3a} \right)\]
\[5 = 4a + 6a\]
\[5 = 10a\]
By simplifying, we get
\[a = 1/2\]
Now put \[a = 2\]in equation \[b = 3a\]
Thus, we get
\[b = 3 \times \dfrac{1}{2}\]
This gives,
\[b = \dfrac{3}{2}\]
Now, put \[a = 1/2;b = 3/2;c = 6;x = 1\] in the equation \[f\left( x \right) = a{x^2} + bx + c\].
So, we get
\[f\left( 1 \right) = \dfrac{1}{2}{\left( 1 \right)^2} + \dfrac{3}{2}\left( 1 \right) + 6\]
\[f\left( 1 \right) = \dfrac{1}{2} + \dfrac{3}{2} + 6\]
By simplifying further, we get
\[f\left( 1 \right) = \dfrac{{1 + 3}}{2} + 6\]
\[f\left( 1 \right) = \dfrac{4}{2} + 6\]
\[f\left( 1 \right) = 2 + 6\]
\[f\left( 1 \right) = 8\]
Hence, the point at which the graph of the function \[f\left( x \right)\]cuts the ordinate\[x = 1\] is \[\left( {1,8} \right)\].
Therefore, the correct option is (A).

Note: Many students make mistakes in finding the values of a, b and c. If so, then they will get the wrong desired answer. That’s why simplification as well as calculation parts are important to get our required solution.