# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | June | Q#7

Question

The point P(4,–1) lies on the curve C with equation y = f(x), x > 0, and

a.   Find the equation of the tangent to C at the point P, giving your answer in the form y = mx + c,  where m and c are integers.

b.   Find f(x).

Solution

a.

We are required to find the equation of tangent to the curve C at point P(4,-1).

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 P(4,-1) on tangent to the curve at point P.

Next, we need to find slope of tangent 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 P(4,-1) 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 P(4,-1).

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;

For gradient of the curve at point P(4,-1), substitute  in derivative of the equation of the curve.

Therefore;

With coordinates of a point on the tangent P(4,-1) and its slope  in hand, we can write equation of the tangent.

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

b.

We are required to find f(x) and we are given f ′(x).

We are also given that the curve passes through the point P(4,-1).

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 the curve passes through the point P(4,-1).

Therefore, substituting given values of y and x.

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