MHB What are f(0), f(1), and f(-1) in the equation f(x)= 2x^2- 3x+ 7?

  • Thread starter Thread starter mathdad
  • Start date Start date
AI Thread Summary
In the equation f(x) = 2x^2 - 3x + 7, the value of c is determined to be 7 from f(0). The values of a and b are found by solving the equations derived from f(1) = 6 and f(-1) = 12, resulting in a = 2 and b = -3. The function can be confirmed by substituting these values back into the original equation. The calculated values for f(0), f(1), and f(-1) are consistent with the derived coefficients. The discussion emphasizes the satisfaction of solving the math problem.
mathdad
Messages
1,280
Reaction score
0
Given f(x) = ax^2+bx+c, where a, b and c are constants, if f(0)=7,what is the value of c? Given that f(1)=6 and f(-1)=12, find the value of a and b.

My Work:

f(0) = 7

7 = a(0)^2 + b(0) + c

7 = 0 + 0 + c

7 = c

f(1) = 6

6 = a(1)^2 + b(1) + 7

6 = a + b + 7

6 - 7 = a + b

- 1 = a + b...Equation A

f(-1) = 12

12 = a(-1)^2 + b(-1) + 7

12 = a - b + 7

12 - 7 = a - b...Equation B

Equations A and B produce a system of equations.

- 1 = a + b...Equation A
12 - 7 = a - b...Equation B

Solving the system of equations for a and b, I found a to be 2 and b to be -3.

Is this correct?
 
Mathematics news on Phys.org
It's easy to check it yourself, isn't it? You are saying that a= 2, b= -3 and c= 7 so you are saying that f(x)= 2x^2- 3x+ 7. So, using that formula, what are f(0), f(1), and f(-1)?
 
HallsofIvy said:
It's easy to check it yourself, isn't it? You are saying that a= 2, b= -3 and c= 7 so you are saying that f(x)= 2x^2- 3x+ 7. So, using that formula, what are f(0), f(1), and f(-1)?

Ok.

- - - Updated - - -

HallsofIvy said:
It's easy to check it yourself, isn't it? You are saying that a= 2, b= -3 and c= 7 so you are saying that f(x)= 2x^2- 3x+ 7. So, using that formula, what are f(0), f(1), and f(-1)?

Ok. It feels good to solve a math problem.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top