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The graph f(x) = ax^2 + bx + c has a gradient of 12 at (4, 21) and a stationary point at x=1. Find the values of a, b and c.
Please help me with this question. Any suggestions would be greatly appreciated. If you have any suggestions, please explain and illustrate explicitly so I understand. Thank you.
Original post by utah
The graph f(x) = ax^2 + bx + c has a gradient of 12 at (4, 21) and a stationary point at x=1. Find the values of a, b and c.
Please help me with this question. Any suggestions would be greatly appreciated. If you have any suggestions, please explain and illustrate explicitly so I understand. Thank you.


A good place to start would be differentiating f(x).
Original post by utah
The graph f(x) = ax^2 + bx + c has a gradient of 12 at (4, 21) and a stationary point at x=1. Find the values of a, b and c.
Please help me with this question. Any suggestions would be greatly appreciated. If you have any suggestions, please explain and illustrate explicitly so I understand. Thank you.


Substitute x=4 into the equation, which will equal 21. Then differentiate f(x), substitute x=1 and make this =0. You'll have two simultaneous equations and should be able to work out the rest from there.
Original post by whatacrydonnie
Substitute x=4 into the equation, which will equal 21. Then differentiate f(x), substitute x=1 and make this =0. You'll have two simultaneous equations and should be able to work out the rest from there.


Following your suggestion I get:

Substituting x=4 into original equation and making it equal to 21
f(x) = ax^2 + bx + c
21 = 16a + 4b + c
0 = 16a +4b + c - 21

Differentiation of f(x), substitution of x=1 and making it equal 0
f(x) = ax^2 + bx + c
f ' (x) = 2ax + b
0 = 2a + b

It cannot be solved simultaneously because of the third variable c. Are you suggesting to ignore c or am I correct?
Original post by utah
Following your suggestion I get:

Substituting x=4 into original equation and making it equal to 21
f(x) = ax^2 + bx + c
21 = 16a + 4b + c
0 = 16a +4b + c - 21

Differentiation of f(x), substitution of x=1 and making it equal 0
f(x) = ax^2 + bx + c
f ' (x) = 2ax + b
0 = 2a + b

It cannot be solved simultaneously because of the third variable c. Are you suggesting to ignore c or am I correct?


remember that dydxx=4=12\dfrac{\text{d} y}{\text {d} x} \Bigg|_{x=4} = 12
Original post by Arithmeticae
remember that dydxx=4=12\dfrac{\text{d} y}{\text {d} x} \Bigg|_{x=4} = 12


Sorry I don't understand, please elaborate.
Original post by utah
Following your suggestion I get:

Substituting x=4 into original equation and making it equal to 21
f(x) = ax^2 + bx + c
21 = 16a + 4b + c
0 = 16a +4b + c - 21

Differentiation of f(x), substitution of x=1 and making it equal 0
f(x) = ax^2 + bx + c
f ' (x) = 2ax + b
0 = 2a + b

It cannot be solved simultaneously because of the third variable c. Are you suggesting to ignore c or am I correct?


Oops sorry, I missed a step.

Differentiate f(x) to get f'(x)= 2ax + b
Substitute x=4, which =12
Substitute x=1 to get 2a + b =0
These are the simultaneous equations which you use to get the value of a and b.
Into f(x) substitute x=4 and f(x)=21 to get the value of c

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Original post by utah
Sorry I don't understand, please elaborate.


the value of the derivative at x=4 is 12, which gives you another equation and allows all the constants to be calculated :smile:
Original post by whatacrydonnie
Oops sorry, I missed a step.

Differentiate f(x) to get f'(x)= 2ax + b
Substitute x=4, which =12
Substitute x=1 to get 2a + b =0
These are the simultaneous equations which you use to get the value of a and b.
Into f(x) substitute x=4 and f(x)=21 to get the value of c

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Cheers bro!
Btw can you suggest a website or maybe even a book where I can find questions similar to this?
Original post by utah
Cheers bro!
Btw can you suggest a website or maybe even a book where I can find questions similar to this?


No problem! examsolutions.net is pretty good, there are loads of questions from different exam boards

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