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

Hits: 45

Question

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.

Solution

a.

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;  b.

We have found in (a) that; To verify f(-2)=5, we can substitute x=-2 in this equation.     c.

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.    Therefore; 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;          