Question

Given \[f\left( x \right) = 12{x^2} + x - 2\] , how do you find the axis of symmetry, vertex, max or min, y-intercept, end behaviour, domain range?

Answer 1

Hint: The given function is of the form, \[a{x^2} + bx + c\] , hence to determine axis of symmetry, apply the vertex formula \[{x_{vertex}} = - \dfrac{b}{{2a}}\] and then to find the vertex substitute the value of \[{x_{vertex}}\] in the given function and to find max or min we must consider the value of a, to find the y-intercept substitute \[x = 0\] and solve for y.
Formula used:
 \[{x_{vertex}} = - \dfrac{b}{{2a}}\] ,
from the Quadratic equation \[\left( {a{x^2} + bx + c} \right)\] .

Complete step by step solution:
Let us write the given function:
 \[f\left( x \right) = 12{x^2} + x - 2\] …………………………. 1
To determine axis of symmetry \[ \to \] vertex
Given, function \[f\left( x \right) = 12{x^2} + x - 2\] is of the form \[a{x^2} + bx + c\] , and hence, to find the axis of symmetry we have \[a = 12\] , \[b = 1\] , \[c = - 2\] , hence we get:
 \[{x_{vertex}} = - \dfrac{b}{{2a}}\]
Substitute the value of a and b from the given function as:
 \[{x_{vertex}} = - \dfrac{1}{{2\left( {12} \right)}}\]
 \[ \Rightarrow {x_{vertex}} = - \dfrac{1}{{24}}\]
Therefore, the axis of symmetry \[ \to x = - \dfrac{1}{{24}}\]
Now, we need to determine Vertex
Hence, substitute \[x = - \dfrac{1}{{24}}\] in the given function of equation 1 as:
 \[{y_{vertex}} = 12{x^2} + x - 2\]
 \[{y_{vertex}} = 12{\left( { - \dfrac{1}{{24}}} \right)^2} + \left( { - \dfrac{1}{{24}}} \right) - 2\]
Simplify the terms, we get:
 \[{y_{vertex}} = 12\left( {\dfrac{1}{{576}}} \right) - \dfrac{1}{{24}} - 2\]
Now, take out the LCM of 576, 24 and 1 as:
 \[{y_{vertex}} = \dfrac{{12}}{{576}} - \dfrac{1}{{24}} - \dfrac{2}{1}\]
Hence, we get LCM as 576, as the GCF is 576; therefore, simplifying the terms we get:
 \[{y_{vertex}} = \dfrac{{12\left( 1 \right)}}{{576}} - \dfrac{{1\left( {24} \right)}}{{24}} - \dfrac{{2\left( {576} \right)}}{1}\]
 \[ \Rightarrow {y_{vertex}} = \dfrac{{12 - 24 - 1152}}{{576}}\]
Evaluating the numerator terms, we get:
 \[ \Rightarrow {y_{vertex}} = - \dfrac{{1164}}{{576}}\]
 \[ \Rightarrow {y_{vertex}} = - \dfrac{{97}}{{48}}or - 2\dfrac{1}{{48}}\]
Therefore, the vertex \[ \to \left( {x,y} \right) = \left( { - \dfrac{1}{{24}}, - \dfrac{{97}}{{48}}} \right)\]
To determine if max or min:
 \[ \to \] Here, in the given quadratic equation, the coefficient of \[{x^2}\] is positive, thus the vertex occurs at a minimum; i.e., if \[a > 0\] , it is a minimum functional value of f.
To determine the y intercept:
 \[ \to \] Substitute \[x = 0\] , to get y-intercept i.e.,
 \[f\left( x \right) = 12{x^2} + x - 2\] , hence
 \[{y_{\operatorname{int} ercept}} = 12\left( 0 \right) + 0 - 2\]
 \[{y_{\operatorname{int} ercept}} = - 2\] ; directly as from the given function \[f\left( x \right)\] .
To determine end behaviour:
 \[ \to \] As x goes on increasing \[12{x^2}\] considerably increases than the rest. As this state grows further the \[ + x - 2\] become insignificant. Thus, we are looking at:
 \[\mathop {\lim }\limits_{x \to \pm \infty } y = \mathop {\lim }\limits_{x \to \pm \infty } 12{x^2} \to 12\left( { \pm \infty } \right) \to \pm \infty \]
To determine domain and range:
 \[ \to \] Range is output: \[f\left( x \right) \to \left[ { - \dfrac{{97}}{{48}},\infty } \right)\]
 \[ \to \] Domain is input of \[f\left( x \right) \to \left\{ {x:x \in R,x \in \left( { - \infty , + \infty } \right)} \right\}\]

Note: We must note that to find maximum or minimum: if \[a > 0\] , it is a minimum functional value of function and if \[a < 0\] , it is a maximum functional value of function. We must know that the domain of a function \[f\left( x \right)\] is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.