0
$\begingroup$

so I'm taking an online Calculus I course and I am currently working through some of the practice problems on the transforming functions section. It's set up to tell me right away if my answer is right or wrong and I have been getting the same type of question wrong over-and-over and was hoping for an explanation since they do not provide one.

The question goes Suppose that y= x^2+4x+1. What is the equation of this function compressed horizontally by a factor of 4? And then it asks me for the new equation.

The way I looked at it was the course taught us that a horizontally compressed function when c>1 is y=f(cx) so applying this to the question I thought it should be y= cx^2+4x(c)+1 am I wrong here?

So evaluated... the answer would be y= (4)x^2 + (4)4x + 1

                             y= 4x^2 + 16x + 1

The reason I am confused is because the course keeps telling me the answer is...

                             y=16x^2 + 16x + 1

But I don't see how we get 16x^2 from an original x^2 when all we are doing is cx^2 when c=4. Is there something I am missing here or is the course answer wrong...I'd love an explanation as simple as this might be.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

You misunderstood the substitution. The substitution is $x\mapsto \color{blue}{cx}$, rather than multiplying the function by $c$. The compressed function should be $$ y= (\color{blue}{4x})^2 + 4(\color{blue}{4x}) + 1 = 16x^2 + 16x + 1 $$ rather than $$ y= (4)x^2 + (4)4x + 1 $$

$\endgroup$
1
  • $\begingroup$ Okay that makes sense. I see what you mean thanks for the help. I wish this course (Outlier.org) if anyone was wondering would should why the solution is what it is. $\endgroup$
    – Noah R.
    Commented Dec 16, 2020 at 3:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .