Parabolas - you are my last resort:
Find a quadratic function in standard form for each set of points. (0, –4), (1, 0), (2, 2)
Can't print answer choices
If it's multiple choice, plug and chug those points.
Otherwise, you have two options:
1) Quadratic Regression
2) Conics
I opt for quadratic regression given that I've not done conics in three years, lol
f(x) = -x^2 + 5x -4
Thank you
To find a quadratic function in standard form, we need to use the general equation of a quadratic function, which is given by:
f(x) = ax^2 + bx + c
where a, b, and c are constants that we need to determine based on the given set of points.
Given the points (0, -4), (1, 0), and (2, 2), we can substitute these coordinates into the equation to form a system of three equations.
Step 1: Substituting (0, -4):
-4 = a(0)^2 + b(0) + c
Simplifying, we get: -4 = c
So, we know that c = -4.
Step 2: Substituting (1, 0):
0 = a(1)^2 + b(1) + (-4)
Simplifying further, we obtain: a + b = 4 (Equation 1)
Step 3: Substituting (2, 2):
2 = a(2)^2 + b(2) + (-4)
Simplifying, we get: 4a + 2b = 6 (Equation 2)
Now, we have a system of two equations with two variables (a and b), which we can solve simultaneously.
Using Equation 1, we can express b in terms of a:
b = 4 - a
Substituting this value into Equation 2, we get:
4a + 2(4 - a) = 6
Simplifying further, we have:
4a + 8 - 2a = 6
2a + 8 = 6
2a = 6 - 8
2a = -2
a = -1
Now that we know the value of a, we can substitute it back into Equation 1 to find b:
(-1) + b = 4
b = 4 + 1
b = 5
So, the values of a, b, and c are: a = -1, b = 5, and c = -4.
Therefore, the quadratic function in standard form that passes through the given points is:
f(x) = -x^2 + 5x - 4.