What is the equation, in standard form, of a parabola that contains the following points?
(-2,-20) (0,-4) (4,-20)
To find the equation of a parabola, we can use the vertex form of a parabola equation, which is given by:
y = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola. To find the values of a, h, and k, we can use the given points (-2, -20) and (0, -4).
Using the point (-2, -20):
-20 = a(-2 - h)^2 + k
And using the point (0, -4):
-4 = a(0 - h)^2 + k
Simplifying the equations:
-20 = a(4 + 4h + h^2) + k
-4 = ah^2 + k
We now have a system of linear equations:
-20 = 4a + 4ah + ah^2 + k
-4 = ah^2 + k
Subtracting the second equation from the first equation:
-20 + 4 = 4a + 4ah + ah^2 + k - ah^2 - k
-16 = 4a + 4ah
-4 = a + ah
Factoring out a:
-4 = a(1 + h)
Now, we can solve for a:
a = -4 / (1 + h)
Substituting this value of a back into one of the original equations (let's use the second equation):
-4 = ah^2 + k
-4 = (-4 / (1 + h))h^2 + k
Simplifying the equation:
-4 = (-4h^2) / (1 + h) + k
-4(1 + h) = -4h^2 + k(1 + h)
-4 - 4h = -4h^2 + k + kh
Simplifying further:
-4 - 4h = -4h^2 + kh + k
0 = -4h^2 + (k - 4h) + k + 4
0 = -4h^2 + (2k - 4h) + 4
Rearranging the equation:
4h^2 - 4h + (2k - 4) = 0
Now, let's use the point (4, -20) to solve for k:
-20 = a(4 - h)^2 + k
-20 = (-4 / (1 + h))(4 - h)^2 + k
Simplifying the equation:
-20 = (-4h^2 + 16h - 16) / (1 + h) + k
-20(1 + h) = -4h^2 + 16h - 16 + k(1 + h)
-20 - 20h = -4h^2 + (16 + k - 4h) + h(k + 1)
Simplifying further:
-20 - 20h = -4h^2 + (h + 16 + k - 4h) + h(k + 1)
-20h = -4h^2 + (1 - 3h + k)
Rearranging the equation:
4h^2 - 20h + (3h - 1 + k) = 0
Now, we have another equation:
4h^2 - 20h + (3h - 1 + k) = 0
Comparing this equation with the previous equation (4h^2 - 4h + (2k - 4) = 0), we can equate the corresponding coefficients:
3h - 1 + k = 2k - 4
Simplifying:
3h - 1 = k - 4
Rearranging the equation:
3h - k = -3
Now we have a system of linear equations:
4h^2 - 4h + (2k - 4) = 0
3h - k = -3
We can now solve this system of linear equations to find the values of h and k.
Solving the second equation for k:
k = 3h + 3
Substituting this value of k into the first equation:
4h^2 - 4h + 2(3h + 3) - 4 = 0
4h^2 - 4h + 6h + 6 - 4 = 0
4h^2 + 2h + 2 = 0
We can now use the quadratic formula to solve for h:
h = (-b ± √(b^2 - 4ac)) / (2a)
h = (-2 ± √(2^2 - 4*4*2)) / (2*4)
h = (-2 ± √(4 - 32)) / 8
h = (-2 ± √(-28)) / 8
Since the square root of a negative number is not real, there are no real solutions for h. This means that the given points (-2, -20), (0, -4), and (4, -20) do not form a parabola.