# What is the slope of the line tangent to $f(x)=-2x^2+4x+6$ when $x=3$?

I'm a 14 year old boy from india, and I'm trying to learn calculus. I've got the following question:

What is the slope of the line tangent to $f(x)=-2x^2+4x+6$ when $x=3$ ?

I've got this formula: $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

I don't understand what $f(x+h)$ is. I only now $f(x)$. Thanks for any help !

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$f(x+h)$ means substituting "$x+h$" wherever $x$ is in $f(x)$. So here $f(x+h) = -2(x+h)^2 + 4(x+h)+6$. – Milind Apr 29 '13 at 16:42
Compute $f'(3)$ – Mathematician Apr 29 '13 at 16:43
Just so we know how much calculus you know so far, do you understand what $\lim_{h\to 0}$ means as far as taking the limit of a function? – Foo Barrigno Apr 29 '13 at 16:44
You guys are so awesome ! So you just place $(x+h)$ everywhere where you see $x$. That is quite simple! :) – user75045 Apr 29 '13 at 16:47
@user75045 : yes, that's exactly how you do it :P – Patrick Da Silva May 11 '13 at 3:39

If $f(x) = -2x^2 + 4x+6$, then $f(y) = -2y^2 + 4y + 6$, $f(z) = -2z^2 + 4z + 6$, and so on. It's just a symbol! So $f(x+h) = -2(x+h)^2 + 4(x+h) + 6$, so if you want to compute the slope of this tangent line, you compute $$\lim_{h \to 0} \frac{( -2(x+h)^2 + 4(x+h) + 6 ) - (-2x^2 + 4x + 6) }h.$$ I'll let you try again! Tell me how it goes.

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Thnank you ! I get : $\lim_{h \to 0} -4x -4h +4$ Then $-4h=0$ ? So $-4x +4$. That is $-12+4=-8$. – user75045 Apr 29 '13 at 16:50
That's correct! – Dolma Apr 29 '13 at 16:51
Thanks all of you ! – user75045 Apr 29 '13 at 17:00
@user75045 : Your answer is correct, but something in there is not! Sorry I had exams so I left MSE alone for a few days. You don't say "$-4h=0$" because you don't want to plug in zero, you know it's not supposed to work! (Which is the whole point of computing the derivative, you don't wanna compute the slope of the line going twice through the same point, that could be anything!) You want to know what happens if $h$ gets as close to zero as it can ever get, so if $h$ goes to $0$, $-4h$ also does. This is what the limit concept is all about. – Patrick Da Silva May 11 '13 at 3:37

Well, since you have $f(x)=-2x^2+4x+6$, then:

$$f(x+h)=-2(x+h)^2+4(x+h)+6$$

The steps are:

1. Expand this equation
2. Remove $f(x)$ from what you expanded
3. Divide it by $h$
4. Determine the limit when $h\rightarrow0$

Note: Another way to solve this problem would be to directly use the derivative. If you've heard about it and if you're interested, then what you're looking for is actually equal to $f'(3)$

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Of course he knows it's "the derivative" in some way ; he's asking for the slope of the tangent line and he has the formula! If he doesn't know derivatives formulas it's probably because he's beginning to learn, so "another way to solve this problem" just means he hasn't "computed this limit yet in general, so he's going at it with an example" for some reason (following a teacher, a book, whatever). :P – Patrick Da Silva May 11 '13 at 3:38