A quadratic function is graphed on a coordinate plane. The parabola passes through the points left parenthesis negative 4 comma 3 right parenthesis, left parenthesis negative 2 comma negative 1 right parenthesis, and left parenthesis 0 comma 3 right parenthesis.
Which equation could be solved using the graph above?
The equation that could be solved using the given graph is a quadratic equation.
The general form of a quadratic equation is y = ax^2 + bx + c, where a, b, and c are constants.
To determine the equation, we will substitute the x and y coordinates of the three points into the equation and solve for a, b, and c.
Using the point (-4, 3):
3 = a*(-4)^2 + b*(-4) + c
3 = 16a - 4b + c
Using the point (-2, -1):
-1 = a*(-2)^2 + b*(-2) + c
-1 = 4a - 2b + c
Using the point (0, 3):
3 = a*(0)^2 + b*(0) + c
3 = c
So we have the system of equations:
3 = 16a - 4b + c ...(1)
-1 = 4a - 2b + c ...(2)
3 = c ...(3)
From equation (3), we can see that c = 3.
Substituting c = 3 into equations (1) and (2), we have:
3 = 16a - 4b + 3 => 16a - 4b = 0 ...(4)
-1 = 4a - 2b + 3 => 4a - 2b = -4 ...(5)
We have a system of linear equations (4) and (5) with two variables a and b. We can solve this system to find the values of a and b.
By solving equations (4) and (5), we get a = 0.25 and b = -0.75.
Therefore, the quadratic equation that could be solved using the given graph is:
y = 0.25x^2 - 0.75x + 3.