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Question: How do you find \[f\left( 6 \right)\] when \[f\left( x \right) = - \dfrac{2}{3}{x^2} - x + 5\]?...

How do you find f(6)f\left( 6 \right) when f(x)=−23x2−x+5f\left( x \right) = - \dfrac{2}{3}{x^2} - x + 5?

Explanation

Solution

In this question we have to find the value of the expression when x=6x = 6, so we input the given value inside the given polynomial, thus substitute the value 6 in place of xx in the polynomial given in the question and further simplification using the operations addition, subtraction and multiplication we will get the required value.

Complete step-by-step answer:
A polynomial is defined as an expression which is composed of variables, constants and exponents that are combined using the mathematical operations such as addition, subtraction, multiplication and division.
Given polynomial is f(x)=−23x2−x+5f\left( x \right) = - \dfrac{2}{3}{x^2} - x + 5,
As the degree of the polynomial is 2 so, the polynomial is a quadratic polynomial.
We have to find the value of f(6)f\left( 6 \right), so substitute the value 6 in place of xx in the given polynomial−23x2−x+5 - \dfrac{2}{3}{x^2} - x + 5, we get,
⇒f(6)=−23(6)2−6+5\Rightarrow f\left( 6 \right) = - \dfrac{2}{3}{\left( 6 \right)^2} - 6 + 5,
Now simplifying the right hand side by taking the square in the polynomial we get,
⇒f(6)=−23(36)−6+5\Rightarrow f\left( 6 \right) = - \dfrac{2}{3}\left( {36} \right) - 6 + 5,
Now further simplification in the right hand side we get,
⇒f(6)=−2(12)−6+5\Rightarrow f\left( 6 \right) = - 2\left( {12} \right) - 6 + 5,
Now taking out the brackets and multiplying the numbers we get,
⇒f(6)=−24−6+5\Rightarrow f\left( 6 \right) = - 24 - 6 + 5,
Now adding for further simplification we get,
⇒f(6)=−30+5\Rightarrow f\left( 6 \right) = - 30 + 5,
Now further simplification we get,
⇒f(6)=−25\Rightarrow f\left( 6 \right) = - 25,
So, the value of the polynomial when x=6x = 6 is -25.

**Final Answer:
∴\therefore The value of the expression f(x)=−23x2−x+5f\left( x \right) = - \dfrac{2}{3}{x^2} - x + 5 when x=6x = 6 will be equal to -25. **

Note:
In solving these questions related to finding the value of a polynomial, we should be careful to ensure that polynomial value doesn’t become indeterminate. For example, if we havef(x)=1x−1f\left( x \right) = \dfrac{1}{{x - 1}}, we cannot find the value of f(1)f\left( 1 \right). Since if we put x=1x = 1, the denominator becomes 0, which is not possible.