Find the quadratic function that satisfies f(0) = 2, f(1) = -3 and f(-1) = 5?

Question

How do you do this problem?

Answer 1

The quadratic function is of the form ax^2 + bx + c = 0.

If f(0) = 2, that tells you that when x equals zero, ax^2 + bx + c = 2. This means c must be 2.

If f(1) = -3, that means a*(1)^2 + b*(1) + 2 = -3.
That tells you a + b = -5

If f(-1) = 5, that means a*(-1)^2 + b*(-1) +2 = 5
That tells you a - b = 3

Add b to both sides of that last equation, and you get a = b+3

Use that to rewrite a + b= -5 as (b+3) + b = -5
Solving this for b, you get b = -4
Since a = b + 3, that means a = -1.

Your quadratic function is: f(x) = -x^2 -4x +2

Answer 2

Any quadratic function can be expressed as

f(x) = a x^2 + b x + c

substituting x = 0 gives

f(0) = 2 = c,

so we have c = 2.

substituting x = 1 gives

f(1) = -3 = a + b + 2

so a = - b - 5.

Substituting x = -1 gives

f(-1) = 5 = a - b + 2

so a = b + 3, and -b -5 = b+3, so b = -4 and a = -1.

f(x) = - x^2 - 4x + 2.

Answer 3

Use the standard form of a quadratic function
f(x) = Ax^2+B*x+C
Then put your given numbers in the equation
For example f(0) = A*0^2+B*0+C = 2
which simplifies to C = 2
Do the same thing with the other two points to get 2 more equations. then solve the system of equations.

Answer 4

f(x) = ax^2 + bx + c

f(x) = ax^2 + bx + 2

f(1) = a + b + 2
f(-1) = a - b + 2

a + b + 2 = -3
a - b + 2 = 5

a + b = -5
a - b = 3

2a = -2
a = -1

a + b = -5
-1 + b = -5
b = -4

ANS : f(x) = -x^2 - 4x + 2

for a graph, just go to http://www.calculator.com/calcs/GCalc.html

type in -(x^2) - 4x + 2 in order to get the graph to show up correctly.