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 Equation of a Straight Line: 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 ... 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 the integral term by term:

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

Combining these results:

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

where $C$ is the constant of integration. To find $C$ , we use the point (4, 25):

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

Calculating:

The first term: $\frac{1}{8}(64) = 8$
The second term: $-5(16) = -80$
The third term: $4$

Thus:

$25 = 8 - 80 + 4 + C \implies 25 = -68 + C \implies C = 93$

So:

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

Step 2

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

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Answer

To find the equation of the normal, we first need the slope of the tangent at the point (4, 25).

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

Thus, the slope of the tangent line is $-33$ . The slope of the normal line is the negative reciprocal: $m_{normal} = -\frac{1}{-33} = \frac{1}{33}$

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (4, 25)$ , the equation becomes: $y - 25 = \frac{1}{33}(x - 4)$
To write it in the form $ax + by + c = 0$ : Multiplying through by 33 to eliminate the fraction: $33(y - 25) = x - 4$ Rearranging: $x - 33y + 839 = 0$

Thus, the integers $a$ , $b$ , and $c$ are: $a = 1, b = -33, c = 839$ .

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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