Graphing new functions based on f(x)

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I have a question: How do you graph new functions based on an original function?

For example, in the following:

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If this graph were to represent the function $f(x)$ then:

  • What would the graph of $-f(-x)$ look like?
  • What would the graph of $f(2x)+3$ look like?
  • What would the graph of $\frac{3}{2}$ $f(-x+1)$

Edit:

Question 3:

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Question 2:

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Question 1:

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3

There are 3 best solutions below

2
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On your graphing calculator, you can graph two different functions at the same time to see the transformation.

For example, if you wanted to graph $f(x)$ and $-f(-x)$ and $f(x)=x^2$, then $-f(-x)=-((-x)^2)=-x^2$. Then you would graph both functions at the same time under the "$Y=$" function button on your calculator.

If you wanted to do this for the other transformations $f(2x)+3$ and $\frac{3}{2}f(-x+1)$, we could find the product of the transformed function and then graph!

1
On

In general, given a graph of a function $f(x)$, it is fairly easy to draw the following things:

$$f(ax)$$ $$f(x+b)$$ $$cf(x)$$ $$f(x)+d$$ And any combination of the above.

Why is this easy? Because $f(x)$ means the value of the function $f$ over all valid values of $x$.

When we do something to $x$ inside the function, we are transforming the domain. Suppose we had our function graphed from $x=0$ to $x=10$. Well, if $x$ ranges from $0$ to $10$, then, $\frac12 x$ ranges from $0$ to $5$. So the graph of $f(\frac12 x)$ from 0 to 10 looks just like the graph of $f(x)$ from 0 to 5, but spread out a bit. Instead of thinking in terms of $x$, pretend $\frac12 x$ is a new, but similar variable, called $x_1$. Plotting $f(\frac12 x)$ is just like plotting $f(x_1)$. Notice that the only difference between the notation $f(x)$ and $f(x_1)$ is that the name of the argument changes. So instead of replacing the function, we're replacing the $x$ axis with the $x_1$ axis. And since $x_1 = \frac12 x$, the $x_1$ axis grows at half the speed as the $x$ axis does.

Likewise $x+b$ shifts the $x$ axis $b$ units to the right (meaning the function shifts $b$ units left).

Similar arguments will show that $cf(x)$ makes the $y$ axis grow at a rate proportional to $c$. If we considered $2f(x)$, then the function is twice as tall as before! If we look at $f(x)+d$, then we shift the $y$ axis down by $d$ units, meaning the function goes up!

It's easy to play with. Simply use some software to plot $f(x) = x^2$, $f(ax) = (ax)^2$, $f(x+b) = (x+b)^2$, $cf(x) = cx^2$ and $f(x)+d = x^2+d$ for whatever values of $a,b,c,d$ you want and see what happens!

2
On

$-f(x)$ means that every positive $f(x)$ (i.e., every positive 'y'-value, or the height of the function) will become negative, and every negative will become positive. Graphically, it means you flip the graph in the x-axis. $f(-x)$ is the same deal, but for your x-values - so you flip it in the x-axis.

For the others, I'd recommend going to desmos.com/calculator and entering $f(x)=x^2$, then $af(bx+c)+d$ and playing around with the sliders to get a sense for the different transformations. That's what my teacher recommended, and it gives you a good geometric intuition, but it's very important to learn the vocabulary to describe what's going on, and that's more of a matter of rote learning.

Also, yes, your graph looks about right. The graph of $-f(-x)$ should resemble a reflection in the line $y=-x$