Write an equation for a parabola with x-intercepts (-1,0) and (4,0) which passes through the point (3,-12)
To find the equation of a parabola, we can start by using the vertex form, which is given by:
y = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
We are given that the x-intercepts are (-1,0) and (4,0), which means the parabola crosses the x-axis at these points. Since the y-coordinate at the x-intercepts is 0, we can substitute these values into the equation:
0 = a(-1 - h)^2 + k
0 = a(4 - h)^2 + k
Next, we are given that the parabola passes through the point (3,-12). We can substitute these values into the equation as well:
-12 = a(3 - h)^2 + k
We now have a system of three equations:
1) 0 = a(-1 - h)^2 + k
2) 0 = a(4 - h)^2 + k
3) -12 = a(3 - h)^2 + k
To solve this system, we can use the elimination method. Subtracting equation 2) from equation 1), we get:
0 = a(-1 - h)^2 + k - a(4 - h)^2 - k
0 = a((-1)^2 + 2(-1)(h) + h^2) - a((4)^2 + 2(4)(h) + h^2)
0 = a + 2ah + ah^2 - a - 8a - 2ah - ah^2
0 = -6a - 2ah
Simplifying further, we have:
-6a - 2ah = 0
-2a(3 + h) = 0
Since a cannot be zero, we have:
h = -3
Substituting h = -3 back into equation 1), we can solve for k:
0 = a(-1 - (-3))^2 + k
0 = a(2)^2 + k
0 = 4a + k
k = -4a
Substituting h = -3 and k = -4a back into equation 3), we can solve for a:
-12 = a(3 - (-3))^2 + k
-12 = a(6)^2 + (-4a)
-12 = 36a - 4a
0 = 32a
a = 0
Since a = 0, the equation of the parabola is:
y = 0(x - (-3))^2 + (-4a)
y = -4a
Thus, the equation of the parabola is y = 0.
To write the equation of a parabola, we need to use the standard form of a quadratic equation: y = ax^2 + bx + c.
Given the x-intercepts of (-1, 0) and (4, 0), we know that the parabola crosses the x-axis at these points. This means that when x = -1 and x = 4, the y-value will be zero.
We can substitute these values into the equation to get two equations:
When x = -1, y = 0:
0 = a(-1)^2 + b(-1) + c
0 = a - b + c
When x = 4, y = 0:
0 = a(4)^2 + b(4) + c
0 = 16a + 4b + c
Additionally, the parabola passes through the point (3, -12). We can substitute these coordinates into the equation:
-12 = a(3)^2 + b(3) + c
-12 = 9a + 3b + c
Now we have a system of three equations:
1) 0 = a - b + c
2) 0 = 16a + 4b + c
3) -12 = 9a + 3b + c
To solve this system, we can use substitution or elimination.
Let's solve it by elimination. First, let's eliminate the variable c by subtracting equation 1) from equation 3):
-12 - (0) = 9a + 3b + c - (a - b + c)
-12 = 8a + 4b
Now, we have two equations remaining:
2a + b = 3 (equation 4)
8a + 4b = -12 (equation 5)
Let's multiply equation 4) by 4 to make the coefficients of b in equations 4) and 5) equal:
8a + 4b = 12 (equation 6)
Now, subtract equation 6) from equation 5):
(8a + 4b) - (8a + 4b) = -12 - 12
0 = -24
This suggests that there is no solution to the system of equations. However, a parabola should always have a solution.
Thus, the given information is inconsistent, and it is not possible to find an equation for a parabola that satisfies the conditions of having x-intercepts at (-1,0) and (4,0), while also passing through the point (3,-12).