- 1). Recall the general form of a quadratic equation: f(x) = ax^2 + bx + c. There are other forms of the quadratic equation, however use this form to find coefficients when given the x and y intercepts.
- 2). Write the given equation as a function of f using the coordinates of the given x intercepts. For example, if given a graph with x intercepts of (-2, 0) and (-1, 0) and a y intercept of (0,4), recall that because this is a quadratic equation and its graph is parabolic there are two places where it hits the x axis. Because the x intercepts in this case are (-2,0) and (-1, 0) they correspond to (x+2) and (x+1). This is because if (x+2) is solved for x the result is x = -2 and if (x+1) is solved for x it yields x = -1. These are the given x intercepts. Plug the values for the x intercepts into the beginning of the equation:
f(x) = a(x+2)(x+1) - 3). Plug the given y intercept into the equation. The given y intercept in this case is (0,4) so 4 takes the place of f(x), or y, and 0 takes the place of every x. Using the figures from the example yields:
f(x) = a(x+2)(x+1)
4 = a(0 + 2)(0 + 1) - 4). Solve the resulting equation for "a." Following the example yields:
4 = a(0+2)(0+1)
4 = a(2)(1)
4 = a2
4/2 = a
a=2 - 5). Plug the value for "a" into the "first step" of the general form of the quadratic equation along with the given intercepts:
f(x) = a[(x+2)(x+1)]
f(x) = 2[(x+2)(x+1)] - 6). Expand the resulting equation to get the general form and the desired coefficients:
f(x) = 2[(x+2)(x+1)]
f(x) = 2(x^2+x+2x+2)
f(x) = 2(x^2 +3x+2)
f(x) = 2x^2+6x+4
a=2, b=6, c=4 - 1). Recall the standard form of the quadratic equation:
f(x) = a(x - h)^2 + k, where the point (h,k) is the vertex.
(a is not equal to 0) - 2). Write the given equation as a function of f using the given y coordinate and the given vertex. If given a graph with a vertex of (-2,-4) and y intercept of (0,4), plug in -2 for h and -4 for k:
f(x) = a(x - k)^2 + k
f(x) = a(x - (-2))^2 + (-4)
f(x) = a(x+2)^2 - 4 - 3). Plug the y intercept into the equation. Remember that the point (0,4) means that the value for x is 0 and the value for y is 4. Since the y intercept is (0,4) write, f(0) = 4. Taking the equation from the example that has the vertices plugged in, plug in the y intercept (0 for x and 4 for y) into the equation:
f(x) = a(x+2)^2 - 4
f(0) = 4, therefore,
4 = a(0+2)^2 - 4 - 4). Solve the equation for "a."
4 = a(0+2)^2 - 4
4 = a(2)^2 - 4
4 = a4 - 4
8 = a4
8/4 = a
a = 2 - 5). Plug the value for "a" into the standard form along with the given intercept and vertex:
f(x) = 2(x+2)^2 - 4 - 6). Expand the resulting equation to get the general form and the desired coefficients:
f(x) = 2(x+2)^2 - 4
f(x) = 2[(x+2)(x+2)] - 4
f(x) = 2(x^2+4x+4) - 4
f(x) = 2x^2+8x+8 - 4
f(x) = 2x^2+8x+4
a=2, b=8, c=4 - 1). Recall the general form of the quadratic equation, f(x) = ax^2 + bx + c, and remember that given three unknowns -- a, b and c -- three points are needed to write three equations to solve for the unknowns.
- 2). Plug the value of the given y intercept into the general form of the equation. If given the x intercepts of (-2,0), (-1, 0) and the y intercept of (0,4), then plugging in the values for the y intercept yields:
f(0) = 4
4 = a(0)^2 + b(0) + c
4 = c - 3). Plug the value of one of the other given points into the equation to solve for the next variable. Given the point (-1,0):
f(-1) = 0 = a(-1)^2 + b(-1) + 4
0 = a - b + 4
a = b - 4 - 4). Plug the values for a and c into original equation to solve for b.
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