# Math Help - secant

1. ## secant

Determine an expression, in simplified form, for the slope of the secant PQ

a) $P(1,3), Q(1+h,f(1+h))$, where $f(x)=3x^2$

Now, i know which formula to use, which is

$Limh->0 \frac{(x+h)-f(x)}{h}$

i have the solution to this problem, but don't understand where the numbers are coming from so it's useless to me. i don't understand how to plug the numbers into this formula

for the first step it is:

$Limh->0 \frac{3(1+h)^2-3}{h}
$

then simply simplifying to get

3h+6

can someone please explain this to me?

2. Originally Posted by euclid2
Determine an expression, in simplified form, for the slope of the secant PQ

a) $P(1,3), Q(1+h,f(1+h))$, where $f(x)=3x^2$

Now, i know which formula to use, which is

$Limh->0 \frac{(x+h)-f(x)}{h}$

i have the solution to this problem, but don't understand where the numbers are coming from so it's useless to me. i don't understand how to plug the numbers into this formula

for the first step it is:

$Limh->0 \frac{3(1+h)^2-3}{h}
$

then simply simplifying to get

3h+6

can someone please explain this to me?
Formula should be $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

You're taking $x=1$ and hence $1+h$.

So $f(1) = 1$ and $f(1+h) = 3(1+h)^2$.

Hence, $\lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \frac{3(1+h)^2 - 3}{h}= \frac{3h^2 + 6h}{h} = 3h + 6 = 6$ as $h \to 0$

3. Originally Posted by Deadstar
Formula should be $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

You're taking $x=1$ and hence $1+h$.

So $f(1) = 1$ and $f(1+h) = 3(1+h)^2$.

Hence, $\lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \frac{3(1+h)^2 - 3}{h}= \frac{3h^2 + 6h}{h} = 3h + 6 = 6$ as $h \to 0$
so what you're doing is plugging (1+h) into 3x^2 to get 3(1+h)^2 ??

and thanks for the help!!

4. Originally Posted by euclid2
so what you're doing is plugging (1+h) into 3x^2 to get 3(1+h)^2 ??

and thanks for the help!!
Yeah, see what I think is going on in the question is you are being asked to find the slope of the curve at the point (1,3).

Now you can do this simply by differentiating f(x) (= 6x) and plugging in x=1.

But it seems they want you to use the formula...

$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Read this to learn more.
Derivative - Wikipedia, the free encyclopedia

Since we are using x=1 (as in P(1,3)).

Plug x into the above formula and just expand it.