A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = rac{3}{8}x^2 - 10x + 1,\, x > 0$ (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 Basic Trigonometry: 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 10 - 2014 - Paper 1

Question 10

A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = rac{3}{8}x^2 - 10x + 1,\, x > 0$ (a) find $f(x)$, simplifying each term... 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 10 - 2014 - Paper 1

Step 1

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

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Answer

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

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

Calculating these integrals, we have:

For the first term:
$\int x^2 \, dx = \frac{x^3}{3}$
Therefore,
$\frac{3}{8} \cdot \frac{x^3}{3} = \frac{x^3}{8}$
For the second term:
$\int x \, dx = \frac{x^2}{2}$
Therefore,
$-10 \cdot \frac{x^2}{2} = -5x^2$
For the constant term:
$\int 1 \, dx = x$

Combining these results, we get:

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

To find $C$ , we utilize the point (4, 25):

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

Calculating the left-hand side:

$= \frac{64}{8} - 80 + 4 + C = 8 - 80 + 4 + C = -68 + C$

Setting this equal to 25 gives:

$-68 + C = 25 \rightarrow C = 93$

Thus,

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

Step 2

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

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Answer

First, we need to find the slope of the tangent line at the point (4, 25) using $f'(x)$ :

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

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

$m_{normal} = -\frac{1}{-33} = \frac{1}{33}$

Using point-slope form, the equation of the normal line at the point (4, 25) is:

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

$y - 25 = \frac{1}{33}x - \frac{4}{33}$
Multiplying through by 33 to eliminate the fraction yields:

$33y - 825 = x - 4 \rightarrow x - 33y + 821 = 0$
Thus, the equation in the form $ax + by + c = 0$ is:

$x - 33y + 821 = 0$
Where $a = 1$ , $b = -33$ , and $c = 821$ .

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 10 - 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 10 - 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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