To find the equation of the quadratic function that passes through the points A(2,3), B(-1,0), and C(1,0), we can use the fact that a quadratic function can be written in the form:
f(x) = ax^2 + bx + c
where a, b, and c are constants. We can set up a system of three equations by plugging in the coordinates of each of the three points:
3 = a(2)^2 + b(2) + c (equation 1, since the point A(2,3) is on the parabola)
0 = a(-1)^2 + b(-1) + c (equation 2, since the point B(-1,0) is on the parabola)
0 = a(1)^2 + b(1) + c (equation 3, since the point C(1,0) is on the parabola)
Simplifying each equation, we get:
4a + 2b + c = 3
a - b + c = 0
a + b + c = 0
We can solve this system of equations by adding the second and third equations, which eliminates b:
2a + 2c = 0
Or simplifying even further:
a + c = 0
This means that we can solve for one of the constants (say, c) in terms of the other (a), substitute into equation 2 to solve for b, and then substitute all three constants into the equation for the quadratic function.
From equation 3, we have:
c = -a
Substituting into equation 2, we get:
0 = a - b - a
Simplifying:
b = 2a
Finally, substituting a, b, and c into the equation for the quadratic function, we get:
f(x) = ax^2 + bx + c
f(x) = ax^2 + 2ax - a
f(x) = a(x^2 + 2x - 1)
To find the value of a, we can use any of the three equations from the beginning. Let's use equation 1, which gives:
3 = a(2)^2 + 2a(2) - a
3 = 4a + 4a - a
3 = 7a
a = 3/7
Substituting this value of a into the equation for the quadratic function, we get:
f(x) = (3/7)(x^2 + 2x - 1)
Now we can graph this function by plotting the points A(2,3), B(-1,0), and C(1,0) and drawing a parabola that passes through them. The vertex of the parabola is at x = -1 and the axis of symmetry is x = -1/2. The graph looks like this:
Find an equation of quadratic function the graph passes through the point A(2,3)B(-1,0) and C (1,0) and draw parabola
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