Write an equation in standard form for the parabola that passes through the points: (-2, 64), (3, -16), (7, 28)
using y = ax^2+bx+c, plug in your points.
4a-2b+c = 64
9a+3b+c = -16
49a+7b = 28
y = 10/3 x^2 + 58/3 x + 12
thank you!
So how would you solve- Find the equation of a parabola that has a vertex of (-3,2) and contains the point (4,7)
To write the equation of a parabola in standard form, we need the coordinates of at least three points on the parabola. Given the points (-2, 64), (3, -16), and (7, 28), we can start by substituting these into the general vertex form equation of a parabola:
y = a(x - h)^2 + k,
where (h, k) represents the vertex. Substituting the coordinates into the equation, we get the following three equations:
64 = a(-2 - h)^2 + k,
-16 = a(3 - h)^2 + k,
28 = a(7 - h)^2 + k.
Now, we can solve this system of equations to find the values of a, h, and k, which will allow us to write the equation in standard form.
First, let's simplify the equations:
64 = a(4 + h)^2 + k,
-16 = a(h - 3)^2 + k,
28 = a(h - 7)^2 + k.
Next, let's expand the squares:
64 = a(16 + 8h + h^2) + k,
-16 = a(h^2 - 6h + 9) + k,
28 = a(h^2 - 14h + 49) + k.
Now, we have a system of three equations with three variables. We can use any method to solve this system, such as substitution or elimination.
One approach is to eliminate k by subtracting the second equation from the first equation and the third equation from the second equation. This will eliminate k and leave us with two equations in two variables:
80 = 14ah - 8ah^2,
44 = 8ah - h^2.
Rearranging the first equation, we get:
5 = ah - \frac{a}{2}h^2.
Now, let's substitute the value of ah from the second equation into the first equation:
5 = \frac{(44 - h^2)}{8}h - \frac{a}{2}h^2,
40 = (44 - h^2) - 4ah^2.
Simplifying:
40 = 44 - h^2 - 4ah^2,
h^2 + 4ah^2 = 4.
Adding the two equations:
h^2 + 4ah^2 + 5 = 4 + 5,
5h^2 = 9.
Dividing both sides by 5:
h^2 = \frac{9}{5}.
Taking the square root of both sides:
h = \pm\sqrt{\frac{9}{5}}.
Next, we can substitute the value of h into the second equation to solve for a:
44 = 8a\sqrt{\frac{9}{5}} - \frac{9}{5}.
Simplifying:
\frac{9}{5} = 8a\sqrt{\frac{9}{5}},
a\sqrt{\frac{9}{5}} = \frac{9}{40},
a = \frac{9}{40}\sqrt{\frac{5}{9}},
a = \frac{3}{40}\sqrt{5}.
Now that we have the values of a and h, we can substitute them into one of the original equations. Let's choose the first equation:
64 = a(-2 - h)^2 + k.
Substituting the values:
64 = \frac{3}{40}\sqrt{5}(-2 - \sqrt{\frac{9}{5}})^2 + k.
Simplifying:
64 = \frac{3}{40}\sqrt{5}(-2 - \frac{9}{5})^2 + k,
64 = \frac{3}{40}\sqrt{5}(-\frac{19}{5})^2 + k,
64 = \frac{3}{40}\sqrt{5}(\frac{361}{25}) + k,
64 = \frac{361\sqrt{5}}{200} + k,
k = 64 - \frac{361\sqrt{5}}{200},
k = \frac{12800 - 361\sqrt{5}}{200},
k = \frac{3200 - 91\sqrt{5}}{50}.
Finally, substituting the values of a, h, and k into the equation of a parabola in standard form, we have:
y = \frac{3}{40}\sqrt{5}\left(x - \sqrt{\frac{9}{5}}\right)^2 + \frac{3200 - 91\sqrt{5}}{50}.
Therefore, the equation in standard form for the parabola passing through the given points is:
y = \frac{3}{40}\sqrt{5}\left(x - \sqrt{\frac{9}{5}}\right)^2 + \frac{3200 - 91\sqrt{5}}{50}.