# Homework Help: Gradient of the secant

1. Jan 6, 2010

### zebra1707

1. The problem statement, all variables and given/known data

There are two parts to this problem.

On the curve 2x^2-5 lie two points P and Q. Let the abscissa of P be "x" and the abscissa of Q be "x+h". No numerical coordinates are given.

a) State the coordinates of P and Q.

b) Using these points find the gradient of the secant PQ

2. Relevant equations

3. The attempt at a solution

For the first part of the question Im assuming that the coordinates would be P (x, f(x)) and Q (x+h, f(x+h))

Now when it comes to finding the gradient of the secant - Can use a table to find a set of coordinates and then plug in the m = y2 - y1 / x2 - x1 ? Not should why the question would ask to use the above coordinates when the gradient of the secant is a different equation?

Could use some guidence on how I attack this problem. Many thanks.

2. Jan 6, 2010

### Dick

Just evaluate the coordinates of P and Q in terms of x and h and using the explicit function f(x)=2x^2-5. Find a formula for the gradient of the secant. Nothing else to do.

3. Jan 6, 2010

### zebra1707

Okay - correct me here if necessary

f(x) = 2x^2 - 5
f(x+h) = 2(x+h)^2 - 5
= 2x^2+4xh+2h^2 - 5

f(x+h) - f(x) = 4xh+2h^2

f(x+h)-f(x) / h = 4x + 2h

f'(x) lim = 4x+2h
approach 0 = 4x gradient of the secant is 4x

Can you please take a moment to check my calculation to confirm my understanding.

4. Jan 6, 2010

### Dick

The calculation is fine. But what you have found is that the gradient of the secant is 4x+2h. Yes, if you take the limit h->0 you get 4x, but that's not the gradient of the secant. That's the gradient of the TANGENT at x. You are jumping ahead.

5. Jan 6, 2010

### zebra1707

Many thanks. So, 4x+2h is the gradient of the secant and I should leave the statement there?

6. Jan 6, 2010

Sure.

7. Jan 6, 2010

### zebra1707

Many thanks Dick - this is greatly appreciated.

Cheers