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 Simultaneous Equations: 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 6 - 2014 - Paper 1

Question 6

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 ... 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 6 - 2014 - Paper 1

Step 1

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

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Answer

To find $f(x)$ , we start by integrating the derivative $f'(x)$ :

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

Now, we will integrate each term separately:

For the first term, ( \int \frac{3}{8}x^2 , dx = \frac{3}{8} \cdot \frac{x^3}{3} = \frac{1}{8}x^3 ).
For the second term, ( \int -10x , dx = -10 \cdot \frac{x^2}{2} = -5x^2 ).
For the last term, ( \int 1 , dx = x ).

Thus, after integrating:

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

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

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

Calculating:

$\frac{1}{8}(64) - 80 + 4 + C = 25 \Rightarrow 8 - 80 + 4 + C = 25 \Rightarrow C = 25 + 72 = 97$

Therefore, the function $f(x)$ is:

$f(x) = \frac{1}{8}x^3 - 5x^2 + x + 97$

Step 2

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

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Answer

To find the equation of the normal to the curve at the point (4, 25), we first need the slope of the tangent line at that point:

Calculate $f'(4)$ :

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

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

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

Now, using the point-slope form of the equation of a line, we can write:

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

Rearranging gives:

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

Multiplying through by 33 to eliminate the fraction:

$33y - 825 = x - 4$

Putting this in the standard form $ax + by + c = 0$ :

$-x + 33y + 821 = 0$

Thus, letting $a = -1$ , $b = 33$ , and $c = 821$ , the final equation of the normal is:

$-x + 33y + 821 = 0$

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 6 - 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 6 - 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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