# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2005 | January | Q#7

**Question**

The curve C has equation , . The point P on C has x-coordinate 1.

**a) **Show that the value of at P is 3.

**b) **Find an equation of the tangent to C at P.

This tangent meets the x-axis at the point (k,0).

**c) **Find the value of k.

**Solution**

**a) **

We need to find the gradient of the curve C at point P(1,y).

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 equation of the curve as;

Therefore;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

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

**b) **

We are required to find the equation of tangent to the curve C at point P(1,0).

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 x-coordinate of a point on the tangent as P(1,y).

With x-coordinate of point of on the curve at hand, we can find the y-coordinate of the point by substituting value of x-coordinate of the point in equation of the curve.

We are given equation of the curve as;

Substituting ;

Hence coordinates of point P on the curve C are .

We need to find slope of tangent at 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(1,8) then we can find slope of the tangent to the curve at this point.

We already have found in (a) gradient of the curve C at point P(1,8).

Therefore;

With coordinates of a point on the tangent P(1,8) and its slope in hand, we can write equation of the tangent.

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

**c)
**

Since tangent meets the x-axis at point (k,0), this is x-intercept of the tangent.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

We have found equation of tangent in (b) as;

Therefore, substituting and ;

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