A-Level Edexcel Maths Pure Simultaneous Equations: The curve C has equation $y = f(

The curve C has equation $y = f(x), x > 0$, where \[f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}.\] Given that the point P(4, -8) lies on C, (a) find the equation of the tangent to C at P, giving your answer in the form $y = mx + c$, where m and c are constants - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 1

Question 8

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The curve C has equation $y = f(x), x > 0$, where \[f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}.\] Given that the point P(4, -8) lies on C, (a) find the equation of the... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = f(x), x > 0$, where \[f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}.\] Given that the point P(4, -8) lies on C, (a) find the equation of the tangent to C at P, giving your answer in the form $y = mx + c$, where m and c are constants - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 1

Step 1

find the equation of the tangent to C at P

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Answer

To find the equation of the tangent, we need to first determine the derivative at the given point.

Substitute $x = 4$ into $f'(x)$ to find the gradient:

$f'(4) = 30 + \frac{6 - 5 \cdot 4^2}{\sqrt{4}} = 30 + \frac{6 - 80}{2} = 30 - 37 = -7.$

This gives us the slope (gradient) of the tangent line, $m = -7$ .

Next, we can use the point-slope formula for the tangent line:

$y - y_1 = m(x - x_1),$ where $(x_1, y_1) = (4, -8)$ .

Substituting the values:

$y - (-8) = -7(x - 4).$

Simplifying this gives:

$y + 8 = -7x + 28$ $y = -7x + 20.$

Thus, the equation of the tangent to C at P is:

$y = -7x + 20.$

Step 2

Find f(x), giving each term in its simplest form

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Answer

To find $f(x)$ , we integrate $f'(x)$ :

Start with the given derivative:

$f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}.$

We can rewrite this as:

$f'(x) = 30 + 6x^{-1/2} - 5x^{3/2}.$
Now, integrate $f'(x)$ :

$f(x) = \int(30 + 6x^{-1/2} - 5x^{3/2}) \,dx$

This results in:

$f(x) = 30x + 6 \cdot 2x^{1/2} - \frac{5}{\frac{5}{2}} x^{5/2} + C$

Simplifying gives:

$f(x) = 30x + 12\sqrt{x} - 2x^{5/2} + C.$
To find the constant $C$ , use the point $P(4, -8)$ that lies on C:

$-8 = 30(4) + 12(2) - 2(32) + C$ $-8 = 120 + 24 - 64 + C$ $C = -8 - 80 = -88.$

Thus, the function is:

$f(x) = 30x + 12\sqrt{x} - 2x^{5/2} - 88.$

find the equation of the tangent to C at P

Find f(x), giving each term in its simplest form

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