Question

If $f(x) = {x^2} + 2x + 3$ , then how do you find $f(a - 1)$ ?

Answer 1

Hint: Firstly, write down the function $f(x)$ in terms of $a$ which will be $f(a)$ . In terms of $a$ mean writing $a$ in place of $x$ . Now, write $f(x)$ in terms of $f(a - 1)$ which is writing $a - 1$ in place of $x$ . Simplify to get an expression in terms of $a$ .

Complete step-by-step solution:
The given function is, $f(x) = {x^2} + 2x + 3$
The given expression is a quadratic function in the form of, $f(x) = a{x^2} + bx + c$
Here, $a,b,c\;$ are constant and $x$ is a variable.
With respect to our given function,
$\Rightarrow a = 1;b = 2;c = 3$ and $x$ is the only variable.
Now, let’s write $f(a)$
To write $f(a)$ we have to substitute $a$ in place of $x$ .
It is a polynomial function that’s why we almost get the same expression but in terms of $a$ .
$\Rightarrow f(x) = {x^2} + 2x + 3$
$\Rightarrow f(a) = {a^2} + 2a + 3$
In the same way, write $f(x)$ as $f(a - 1)$
To write $f(a - 1)$ we have to substitute $a - 1$ in place of $x$ .
$\Rightarrow f(x) = {x^2} + 2x + 3$
$\Rightarrow f(a - 1) = {(a - 1)^2} + 2(a - 1) + 3$
Now start simplifying the expression.
Open the squared term, write it twice and then multiply.
$\Rightarrow f(a - 1) = [(a - 1)(a - 1)] + 2(a - 1) + 3$
$\Rightarrow f(a - 1) = [a(a - 1) - 1(a - 1)] + 2(a - 1) + 3$
Now open the brackets and simplify
Also, add the constants.
$\Rightarrow f(a - 1) = [({a^2} - a) - (a - 1)] + 2a - 2 + 3$
$\Rightarrow f(a - 1) = [{a^2} - a - a + 1] + 2a + 1$
Add the same degree terms.
$\Rightarrow f(a - 1) = {a^2} - 2a + 1 + 2a + 1$
Now, add the constants.
$\Rightarrow f(a - 1) = {a^2} + 1 + 1$
Evaluate further.
$\Rightarrow f(a - 1) = {a^2} + 2$

When $f(x) = {x^2} + 2x + 3$ , then $f(a - 1) = {a^2} + 2$ .

Additional information: The function defines a property or a relation between the input and the output such that each input relates to exactly one output. A quadratic function is a function where there is a second-degree polynomial and can be shown in the format, $f(x) = p{x^2} + qx + r$ where $p \ne 0$ and $p,q,r\;$ are constants and $x$ is a variable.

Note: One should be careful while substituting the values in the expression. Whenever a function $f(x)$ is given and we have to find the value of $f(p)$ , just substitute the value of $p$ in place of $x$ in the function to get $f(p)$ .