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The point P (4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 (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 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

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The-point-P-(4,--1)-lies-on-the-curve-C-with-equation-y-=-f(x),-x->-0,-and---f'(x)-=-\frac{1}{2}---\frac{6}{\sqrt{x}}-+-3--(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-Edexcel-A-Level Maths Pure-Question 8-2012-Paper 2.png

The point P (4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 (a) Find the equation of the tangent to C ... show full transcript

Worked Solution & Example Answer:The point P (4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 (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 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

Step 1

Find the equation of the tangent to C at the point P

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Answer

To find the equation of the tangent, we first need to calculate the derivative at the point P. We have:

  1. Calculate f'(4):

f'(4) = \frac{1}{2} - \frac{6}{\sqrt{4}} + 3
= \frac{1}{2} - \frac{6}{2} + 3
= \frac{1}{2} - 3 + 3
= \frac{1}{2}

  1. The slope m of the tangent at point P is therefore ( m = \frac{1}{2} ).

  2. Now we use the point-slope form of a line equation:
    y - y_1 = m(x - x_1)
    where ( P(4, -1) ) gives us ( y_1 = -1 ) and ( x_1 = 4 ).

  3. Substituting the values: y - (-1) = \frac{1}{2}(x - 4)
    y + 1 = \frac{1}{2}x - 2
    y = \frac{1}{2}x - 3

  4. To express in the form y = mx + c with integers, we write:
    y = \frac{1}{2}x - 3 or equivalently:
    2y = x - 6
    which simplifies to y = \frac{1}{2}x - 3, where m = 1 and c = -3.

Step 2

Find f(x)

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Answer

Given: f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3
We need to integrate f'(x) to find f(x):

f(x) = \int f'(x) , dx
= \int \left( \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 \right) , dx
= \frac{1}{2} x - 6 \int x^{-\frac{1}{2}} , dx + 3x + c
= \frac{1}{2} x - 12 \sqrt{x} + 3x + c
= \frac{1}{2} x + 3x - 12 \sqrt{x} + c

Now, substituting point P(4, -1) to find c: f(4) = \frac{1}{2}(4) + 3(4) - 12 \sqrt{4} + c
-1 = 2 + 12 - 24 + c
c = -1 - 2 - 12 + 24
c = 7.

Thus, the function is: f(x) = \frac{1}{2}x + 3x - 12\sqrt{x} + 7.

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