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A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 1
Question 3
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 3 - 2014 - Paper 1
Step 1
find $f(x)$, simplifying each term.
Answer
To find , we need to integrate .
egin{align*}
f'(x) &= rac{3}{8}x^2 - 10x + 1 \
ext{Integrating:} \
f(x) &= rac{3}{8} \cdot rac{x^3}{3} - 10 \cdot rac{x^2}{2} + x + C \
&= rac{1}{8}x^3 - 5x^2 + x + C
ext{Using the point (4, 25) to find } C:\
25 &= rac{1}{8}(4)^3 - 5(4)^2 + 4 + C
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25 &= rac{1}{8}(64) - 80 + 4 + C
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25 &= 8 - 80 + 4 + C
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C &= 25 - 8 + 80 - 4 \
C &= 93
ext{Thus, } f(x) = \
\frac{1}{8}x^3 - 5x^2 + x + 93.
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Step 2
Find an equation of the normal to the curve at the point (4, 25).
Answer
To find the equation of the normal, we first need to determine the slope of the tangent at (4, 25).
Calculating :
egin{align*}
f'(4) &= rac{3}{8}(4)^2 - 10(4) + 1 \
&= rac{3}{8}(16) - 40 + 1 \
&= 6 - 40 + 1 \
&= -33.
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ext{The slope of the normal is the negative reciprocal:} \
m_{ ext{normal}} = - rac{1}{-33} = rac{1}{33}.
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ext{Using point-slope form for the equation of the normal:}
y - 25 = rac{1}{33}(x - 4)
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Rearranging to the form :
33y - 825 = x - 4 \
x - 33y + 821 = 0.\
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