What is the equation, in standard form, of a parabola that contains the following points?
(-1, 10) , (0, 5) , (2, 13)
To find the equation of the parabola in standard form, we need to use the vertex form of the equation of a parabola:
y = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
Let's start by finding the vertex of the parabola using the given points.
The x-coordinate of the vertex (h) can be calculated as the average of the x-coordinates of the given points:
h = (-1 + 0 + 2)/3 = 1/3
The y-coordinate of the vertex (k) can be calculated as the average of the y-coordinates of the given points:
k = (10 + 5 + 13)/3 = 28/3
So, the vertex of the parabola is (1/3, 28/3).
Now, let's substitute the vertex's coordinates into the vertex form equation:
y = a(x - 1/3)^2 + 28/3
Next, let's use one of the given points, (0, 5), to find the value of a.
5 = a(0 - 1/3)^2 + 28/3
5 = a(1/9) + 28/3
5 = a/9 + 28/3
15 = a + 84/9
15 = a + 28/3
15 = a + 56/6
15 = a + 14/2
15 = a + 7
a = 8
Now that we have the value of a, we can write the equation of the parabola in standard form by substituting it into the vertex form equation:
y = 8(x - 1/3)^2 + 28/3
Multiplying through to remove any fractions:
3y = 8(x - 1/3)^2 + 28
3y = 8(x^2 - 2/3x + 1/9) + 28
3y = 8x^2 - 16/3x + 8/9 + 28
3y = 8x^2 - 16/3x + 244/9
Finally, rearrange the equation to standard form:
8x^2 - 16/3x + 244/9 - 3y = 0
Thus, the equation, in standard form, of the parabola that contains the given points is:
8x^2 - 16/3x + 244/9 - 3y = 0