Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2006 | June | Q#10

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The curve C with equation y=f(x), x ≠ 0, passes through the point .

Given that

a.   find f(x).

b.   Verify that f(–2) = 5.

c.   Find an equation for the tangent to C at the point (–2, 5), giving your answer in the form ax + by +  c = 0, where a, b and c are integers.



We are required to find f(x), when;

We are also given that , and the curve passes through the point .

Clearly it is the case of finding equation from its derivative.

We can find equation of the curve from its derivative through integration;

For the given case;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and    in the equation obtained from integration of the derivative of the curve i.e. .

We are also given that , and the curve passes through the point .

Therefore, substituting given values of y and x.

Hence, above equation obtained from integration can now be written as;


We have found in (a) that;

To verify f(-2)=5, we can substitute x=-2 in this equation.


We are required to find the equation of tangent to the curve C at point (-2,5).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We already have coordinates of a point on the tangent as P(-2,5).

We have found equation of the curve in (a) as;

We need to find slope of tangent at (-2,5) in order to write its equation.

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, if we can find slope of the curve C at point (-2,5) then we can find slope of the tangent to  the curve at this point.

We need to find the gradient of the curve C at point (-2,5).

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We are given expression for gradient of the given curve as;

For gradient of the curve at point (-2,5), substitute  in derivative of the equation of the curve.


With coordinates of a point on the tangent (-2,5) and its slope  in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;