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.