Find a parabola, passing through the origin, whose tangent line at (2,4) passes through the point (0,8).

Question

anonymous · Answer

ok if it passes through the origin you know that 
$$f(x)=ax^2+bx=c$$ and 
$$f(0)=c=0$$ so right away you know the constant is 0

anonymous · Answer

otherwise it cannot pass through the origin. constant is 0.  now you know that the tangent line at the point (2,4) passes through (0,8)

anonymous · Answer

the slope of that line is 
$$\frac{8-4}{0-2}=\frac{4}{-2}=-2$$

anonymous · Answer

this means that the derivative at the point (2,4) is -2

anonymous · Answer

$$f(x)=ax^2+bx$$
$$f'(x)=2ax+b$$
$$f'(2)=4a+b=-2$$
so we know 
$$4a+b=-2$$

anonymous · Answer

and finally we know that (2,4) is on the graph of f, which means 
$$f(2)=a\times 2^2+b\times 2 = 4$$
$$4a+2b=4$$

anonymous · Answer

so we have two equations and two unknowns, namely 
$$4a+b=-2$$ and 
$$4a+2b=4$$ subtract first from second to get b=6 etc

anonymous · Answer

$$4a+6=-2$$
$$4a=-8$$
$$a=-2$$ if my arithmetic is write and your function is 
$$f(x)=-2x^2+6x$$

anonymous · Answer

b isn't -6?

anonymous · Answer

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anonymous · Answer

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anonymous · Answer

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anonymous · Answer

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anonymous · Answer

spiderman?

anonymous · Answer

anonymous · Answer

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anonymous · Answer

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