Find quadratic function with points (0,5), (1,2) and (3,2). Graph the function
If y = ax^2+bx+c, then we can plug in the points and get
0a+0b+c = 5
a+b+c = 2
9a+3b+c = 2
y = x^2-4x+5
Or, noting the symmetry about x=2, we know that
y = a(x-2)^2 + k
y(0) = 5, so
4a+k = 5
y(1) = a+k = 2
so, a=1,k=1
y = (x-2)^2+1
To find the quadratic function that passes through the given points (0,5), (1,2), and (3,2), we can use the general form of a quadratic function:
f(x) = ax^2 + bx + c
Substituting the coordinates of the points into the equation, we can set up a system of equations:
For point (0,5):
f(0) = a(0)^2 + b(0) + c = 5
=> 0 + 0 + c = 5
=> c = 5
For point (1,2):
f(1) = a(1)^2 + b(1) + c = 2
=> a + b + c = 2
=> a + b + 5 = 2
=> a + b = -3
For point (3,2):
f(3) = a(3)^2 + b(3) + c = 2
=> 9a + 3b + 5 = 2
=> 9a + 3b = -3
=> 3a + b = -1
We now have a system of equations:
a + b = -3
3a + b = -1
Solving this system of equations, we can subtract the first equation from the second equation:
(3a + b) - (a + b) = -1 - (-3)
2a = 2
a = 1
Substituting a = 1 into the first equation:
1 + b = -3
b = -4
Now that we have the values for a and b, we can write the quadratic function:
f(x) = x^2 - 4x + 5
To graph the function, plot the given points (0,5), (1,2), and (3,2) on a coordinate plane. Then, connect the points with a smooth curve, as the quadratic function is continuous.