Find the quadratic polynomial whose graph goes through the points (-2,5), (0,5), and (1,11).
f(x)= ______x^2+ ______x+ _______
I don't know why teachers assign these, if I got them often, I would make a voodoo doll
start with
f(0)=5=25a+0b+25c for the point (o,5)and
f(-2)=5=4a -10b+25c for the (-2,5) and
f(1)=11 you do it.
Now you have three equations, three unknowns. There are a number of ways to solve these, I am uncertain which ways you have been taught. I would use gaussian elimination on a calculator (see your calculator handbook)
since f(-2) = f(0), symmetry demands that the vertex be at x = -1. So,
y = a(x+1)^2 + k
Now we have
a+k = 5
4a+k = 11
a = 2
k = 3
so,
y = 2(x+1)^2 + 3
To find the quadratic polynomial that goes through the given points, we need to use the general form of a quadratic polynomial, which is f(x) = ax^2 + bx + c.
The first step is to substitute the given points into the equation. Let's start with the point (-2,5):
5 = a(-2)^2 + b(-2) + c
Now, let's substitute the next point (0,5):
5 = a(0)^2 + b(0) + c
Lastly, substitute the third point (1,11):
11 = a(1)^2 + b(1) + c
Now, we have a system of three equations with three unknowns (a, b, and c). We can solve this system of equations to find their values.
From the second equation, we can see that c = 5.
Substituting c = 5 into the first two equations, we have:
5 = 4a - 2b + 5 -> 0 = 4a - 2b -> 2a = b (Equation 4)
11 = a + b + 5 -> 6 = a + b -> b = 6 - a (Equation 5)
Now, substitute Equation 5 into Equation 4:
2a = 6 - a
Solving for a:
3a = 6 -> a = 2
Substitute a = 2 into Equation 5 to find b:
b = 6 - 2 = 4
So, the values of a, b, and c are: a = 2, b = 4, and c = 5.
Therefore, the quadratic polynomial that goes through the given points is:
f(x) = 2x^2 + 4x + 5