• Problem
  • Detailed Solution
  • Summary Solution

Determine the slope of the tangent to the graph of f(x) at point P(2/f(2)).

a) \(\;f(x)=2.5x^2\)

b) \(\;f(x)=-2x^3+x^2-5x+9\)

Here, we discuss problem b) in detail. Problem a) can be solved accordingly.

Structure of the problem

The assignment equation \(f(x)=-2x^3+x^2-5x+9\) is given. We now have to determine the slope of the tangent to the graph of this function in the point P(2/f(2)). To solve this problem, we need to know a couple of things:

(i) What is a tangent?

(ii) What is the slope of a tangent? (See the same place in the script for the theoretical background.)

(iii) How do you determine the slope of a tangent of a function f(x), i.e. how do you find f'(x), for a given value of x ? (The theory behind this is in block II, learning sequence 2, of the script.)

(iv) How can the tangent of a function f(x) be represented graphically?

Sketching the graph of the function

Provided you know enough about problems (i) through (iv), it makes sense to sketch the graph of our function


This will help you to imagine how the solution must look like. Now please sketch the graph of f(x) for values 0 < x < 3 (since we focus on x = 2, it makes sense to study f in a neighbourhood of that place). To verify your results, you can find the graph of f here.

Remark: Alternatives for sketching graphs you find here.

Now that we know how the tangent looks like, we need to know its slope. Later, we'll see that this is -25. In the diagram you can read an interpretation of this value: if, starting in the point (2/-13), we move one step to the right, the value of the tangent goes down by 25 steps, i.e. you need to make 25 steps downwards to meet the tangent again.

Formal approach

So far, we have made use of graphical analysis alone. By consequence, we couldn't expect to be more precise than our drawing skills allow us to be. To be more precise than that, we need to study the first derivative of f, i.e. we must determine the value of f'(x) for x = 2. This is because f'(x) is nothing else but the slope of the tangent to f(x) at (x/f(x)).

We now put our derivation rules to work: Since \(f(x)=-2x^3+x^2-5x+9\) we have \(f'(x)=-6x^2+2x-5\). To find the slope for x = 2, all we need to do is replace x by 2 in that equation: \(f'(2)=-6 \cdot 2^2+2 \cdot 2-5=-25\).


  • Our calculus show us that to determine f'(x), we don't even need to know the value of f(x) to start with. This is because the differential calculus depends exclusively on the equation of f'(x), not on the equation of f(x) itself.
  • Another graphic approach to solving our problem is this (see here): sketch the graph of the derivative function f'(x), find the point (2 / f'(2)), which in our case is -25. This is the value of f'(2), and this ist he number we have been looking for.
  • Note the difference between this problem and the one presented in exercise 3, series 3: There we have to solve the equation f'(x)=2, because we know the slope of the tangent (=2), but not the value of x. Here we know the value x=2 and have to evaluate the derivative function f'(x) at x=2, i.e. we need to calculate f'(2) to get the slope of the tangent.

Exercise a)



\(f'(2)=5 \cdot 2=10\)

The slope of the tangent to the graph of the function \(f(x)=2.5x^2\) at point P(2/f(2)) is 10.

Exercise b)



\(f'(2)=-6 \cdot 2^2+2 \cdot 2-5=-25\)

The slope of the tangent to the graph of the function \(f(x)=-2x^3+x^2-5x+9\) at point P(2/f(2)) is -25.