# Thread: How To Find The Slope Of A Curve...?

1. ## How To Find The Slope Of A Curve...?

I understand that it's delta y / delta x. I'm a bit confused about these two questions.

Get m at x = 1

#1 y = x^2+3x

My work:
((x^2+3x+h) - x^2 - 3x)/h

h/h = 1

m (slope) = 1?

#2 x^2-4x+3

((x^2-4x+3+h) - (x^2-4x+3))/h
h/h = 1

m (slope) = 1?

2. Got it!

I see where I went wrong..

m=5 for #1

m=-2 for #2

Thanks, guys!

3. Hello Alpha Rock:

Your use of the symbol h leads me to believe that you need to use the limit definition of the derivative to find the slope of the tangent line at x = 1.

You did not properly apply this definition because you simply added h in the numerator.

Let's take it from the top.

$f(x) = x^2 + 3x$

$m = f'(x)$

Here is the limit definition of f'(x).

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

I'll focus first on f(x + h).

This is what we call a composite function. The notation tells us that we need to replace each x with x + h in the expression that defines f(x).

$f(x + h) = (x + h)^2 + 3(x + h)$

$= x^2 + 2xh + h^2 + 3x + 3h$

Therefore, the numerator of the ratio in our limit definition is the following.

$f(x + h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x)$

$= x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x$

Notice that the x^2 and -x^2 combine to make zero. Same goes for 3x and -3x.

$f'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h}$

Now, the only thing that prevents us from evaluating this limit by directly substituting zero for h is the fatal h in the denominator. This problem is easily resolved because each term in the numerator contains at least one factor of h, so we can factor it out, and it will then cancel with the h in the denominator.

$f'(x) = \lim_{h \to 0} \frac{h(2x + h + 3)}{h}$

$f'(x) = \lim_{h \to 0} 2x + h + 3$

NOW, we can use direct substitution to evaluate this limit by letting h go to zero.

$f'(x) = 2x + 0 + 3 = 2x + 3$

This formula for f'(x) gives us the slope of the tangent line at any value of x.

$f'(1) = 2(1) + 3 = 5$

I'll let you give it another go on your second exercise.

Cheers,

~ Mark

4. Thanks, mmm4444bot.

I really appreciate this...

Your post helps clear things up A LOT better. Really.

- AlphaRock

(Now I don't have to bang my head against the wall for this question... )