Question

Find a, b, c when $f\left( x \right)=a{{x}^{2}}+bx+c$ and f (0) = 6, f (2) = 11, f (-3) = 6. Determine the value of f (1).

Answer 1

Consider the first case and substitute x = 0 in the given polynomial $f\left( x \right)=a{{x}^{2}}+bx+c$ and equate it with 6 to form a relation in a, b and c. Similarly, form other two relations in a, b and c using the given information. Solve three linear equations to get the values of a, b and c. Substitute these values in $f\left( x \right)=a{{x}^{2}}+bx+c$ to determine the actual polynomial. Now, substitute x = 1 to find the value of f (1).

Complete step-by-step solution
Here, we have been provided with the polynomial $f\left( x \right)=a{{x}^{2}}+bx+c$ and we have to find the values of a, b, and c with some provided information. So, let us form three linear equations to find the values of these variables.
1. When f (0) = 6: -
Substituting x = 0 in f (x), we get,

& \Rightarrow f\left( 0 \right)=a\times {{0}^{2}}+b\times 0+c \\\ & \Rightarrow 6=c \\\ \end{aligned}$$ $$\Rightarrow c=6$$ - (1) 2\. When f (2) = 11: - Substituting x = 2 in f (x), we get, $$\begin{aligned} & \Rightarrow f\left( 2 \right)=a\times {{2}^{2}}+b\times 2+c \\\ & \Rightarrow 11=4a+2b+c \\\ \end{aligned}$$ $$\Rightarrow 4a+2b+c=11$$ - (2) 3\. When f (-3) = 6: - Substituting x = -3 in f (x), we get: - $$\begin{aligned} & \Rightarrow f\left( -3 \right)=a\times {{\left( -3 \right)}^{2}}+b\times \left( -3 \right)+c \\\ & \Rightarrow 6=9a-3b+c \\\ \end{aligned}$$ $$\Rightarrow 9a-3b+c=6$$ - (3) Now, we have three variables and three equations. So, solving equations (1), (2) and (3), we get, $$\Rightarrow a=\dfrac{1}{2},b=\dfrac{3}{2}$$ and c = 6 So, the actual polynomial can be obtained by substituting the obtained values of a, b and c in the given expression of f (x). So, we have, $$\Rightarrow f\left( x \right)=\dfrac{1}{2}{{x}^{2}}+\dfrac{3}{2}x+6$$ Taking $$\dfrac{1}{2}$$ common from all the terms, we get, $$\Rightarrow f\left( x \right)=\dfrac{1}{2}\left( {{x}^{2}}+3x+12 \right)$$ Now, we have to find the value of f (1) which can be obtained by substituting x = 1 in the above expression of f (x). So, we get, $$\begin{aligned} & \Rightarrow f\left( 1 \right)=\dfrac{1}{2}\left( {{1}^{2}}+3\times 1+12 \right) \\\ & \Rightarrow f\left( 1 \right)=\dfrac{1}{2}\left( 1+3+12 \right) \\\ & \Rightarrow f\left( 1 \right)=\dfrac{1}{2}\times 16 \\\ & \Rightarrow f\left( 1 \right)=8 \\\ \end{aligned}$$ **Note:** One may note that in the above question we were given to find the values of three variables a, b and c, so it was necessary to form three equations. If any of the equations get reduced then we will not be able to solve the question. So, in general, remember that if we have to find the values of ‘n’ given variables then we must need ‘n’ number of equations.