A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = \frac{3}{8} x^2 - 10x + 1, \quad x > 0$ (a) find $f(x)$, simplifying each term. (b) Find an equation of the normal to the curve at the point (4, 25). Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers to be found.

A-Level Edexcel Maths Pure Graphs of Functions: A curve with equation $y = f(x)$

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 1

Question 11

A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = \frac{3}{8} x^2 - 10x + 1, \quad x > 0$ (a) find $f(x)$, simplifying each t... show full transcript

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 1

Step 1

find $f(x)$, simplifying each term.

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Answer

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

$f(x) = \int f'(x) \, dx = \int \left( \frac{3}{8} x^2 - 10x + 1 \right) \, dx$

Calculating this gives:

$f(x) = \frac{3}{8} \cdot \frac{x^3}{3} - 10 \cdot \frac{x^2}{2} + x + C$

This simplifies to:

$f(x) = \frac{1}{8} x^3 - 5x^2 + x + C$

Next, we use the point (4, 25) to find the constant $C$ :

$f(4) = \frac{1}{8} (4)^3 - 5(4)^2 + 4 + C \Rightarrow 25 = \frac{1}{8} (64) - 5(16) + 4 + C$ $25 = 8 - 80 + 4 + C$ $25 = -68 + C$ $C = 93$

Thus,:

$f(x) = \frac{1}{8} x^3 - 5x^2 + x + 93$

Step 2

Find an equation of the normal to the curve at the point (4, 25).

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Answer

First, we need the slope of the tangent at the point (4, 25). We can find this by evaluating $f'(4)$ :

$f'(4) = \frac{3}{8} (4)^2 - 10(4) + 1 = \frac{3}{8} (16) - 40 + 1 = 6 - 40 + 1 = -33$

The slope of the normal line is the negative reciprocal of the slope of the tangent:

$m_{normal} = -\frac{1}{f'(4)} = -\frac{1}{-33} = \frac{1}{33}$

Next, we use the point-slope form of a line to find the equation of the normal line:

$y - 25 = \frac{1}{33}(x - 4)$

Rearranging gives:

$y - 25 = \frac{1}{33}x - \frac{4}{33}$ $33(y - 25) = x - 4$ $x - 33y + 825 + 4 = 0$ $x - 33y + 829 = 0$

Thus, the normal can be expressed as:

$1 x - 33 y + 829 = 0$

where $a = 1$ , $b = -33$ , and $c = 829$ .

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 1

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 1

find $f(x)$, simplifying each term.

Find an equation of the normal to the curve at the point (4, 25).

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