I am little confused with the idea of images of a function How do you find the image of a particular interval with respect to a function?
For example : How do you find the image of the interval $[-1,3]$, under the mapping $f:\Bbb R\to \Bbb R$ given by $f(x)=4x^2-12x$?
On
Because the function is a quadratic than it is a continuous function.So it means that the image will contain all the points between the $[\min,\max]$
$$4x^2-12x=3(x-0)(x-3)$$
The function has a minimum cause $a>0$ and it happens at the average of two roots so this time we have $$\frac{(0+3)}{2} = 1.5$$ and the min is $f(1.5)=-9 $, the maximum can be found going as far as possible from the $x=1.5$ which this time in our interval is $x=-1$ (distance from $x=1.5$ is $2.5$ compared to $1.5$ if $x = 3$) so we have max $$f(-1)= 16$$
So the image will be the interval $[\min,\max]=[-9,16]$
We can confirm that by a graphing calculator.
General advice when trying to solve a math problem is to try to visualize it. So when we want to find the image of a function on a certain interval, the first thing we might try is to plot that function. For your example $$f(x) = 4x^2-12x$$ we get something like this
Now we can conjecture what the answer might be, namely the interval $[-9,16]$. From the plot we can also get some idea of what our strategy to prove this claim might be. We can, for instance, use the concept of the derivative to prove that the minimum of $f$ is attained at $x=\frac{3}{2}$ and is equal to $f\left(\frac{3}{2}\right)=-9$. Now you might use symmetry arguments to show that the maximum on $[-1,3]$ must be attained in $x=-1$ and thus be equal to $f(-1)=16$. Then use continuity of the function to show that every value in between $-9$ and $16$ is attained as well.
Be aware that the plot of the function is useful for conjecturing, but can not be used to prove anything definitively. Every claim you make must be proved rigorously.
Another general strategy when tackling math problems is to try and break it up into smaller problems. The function $f$ can be seen as first sending $x$ to $x^2-3x$ and then multiplying the whole thing by $4$. If we let $g(t)=t^2-3t$ and $h(t)=4t$ we get $$f(x) = h(g(x)).$$ This does not seem to make the problem much easier yet, but I claim that the functions $g$ and $h$ are more easy to handle than $f$. Let's focus on $g$ a little more. We have $$ g(t)=t^2-3t=t(t-3). $$ This expression seems horribly asymmetric, but we might make it symmetric if we could just shift the variable by $\frac{3}{2}$. To do this we split $g$ up into two parts again. First send a variable $x$ to $x-\frac{3}{2}$, call this $k$ such that $k(x)=x-\frac{3}{2}$. Then apply $t\mapsto (t+\frac{3}{2})(t-\frac{3}{2})$ and call this function $m$. We see that $$ g(x) = m(k(x)). $$
Now we can investigate what we have gained from this symmetry. If we multiply the terms in $m$ we get $$ m(x) = x^2-\frac{9}{4}. $$ This function is certainly easier to deal with than $f$. We can split it up one more time into first sending $x$ to $x^2$ and then subtracting $\frac{9}{4}$. If we have done our bookkeeping correctly we see that we have now split up $f$ into the following four maps \begin{align*} f_1:&\ x \mapsto x - \frac{3}{2}\\ f_2:&\ x \mapsto x^2 \\ f_3:&\ x \mapsto x - \frac{9}{4} \\ f_4:&\ x \mapsto 4x. \end{align*} Now we have $$f(x)=f_4\left(f_3\left(f_2\left(f_1\left(x\right)\right)\right)\right)$$ and we can just follow the interval around. We get $$ [-1,3]\stackrel{f_1}{\mapsto} \left[-\frac{5}{2},\frac{3}{2}\right] \stackrel{f_2}{\mapsto} \left[0,\frac{25}{4}\right] \stackrel{f_3}{\mapsto} \left[-\frac{9}{4},\frac{16}{4}\right] \stackrel{f_4}{\mapsto} [-9,16]. $$ The images shown in each step here are hopefully easier to find than the image of $f$, if not obvious.