A parabola has its vertex on the graph of the line y=3x+1 and passes through (1,10). If it is the same size,?
shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas
Thanks!!!
let the parabola have equation
y = 3(x-p)^2 + q
(1,10) lies on it, so
10 = 3(1-p)^2 + q
also the vertex (p,q) lies on y = 3x+1
thus:
q = 3p + 1
sub back into equation above
10 = 3(1-p)^2 + 3p + 1
10 = 3 - 6p + 3p^2 + 3p + 1
3p^2 -3p -6 = 0
p^2 - p - 2 = 0
(p-2)(p+1) = 0
p = 2 or p = -1
if p=2, then q = 3(2) + 1 = 7
and the parabola is y = 3(x - 2)^2 + 7
if p = -1, then q = 3(-1) + 1 = -2
and the parabola is y = 3(x+1)^2 - 2
To find the equation(s) of all possible parabolas that satisfy the given conditions, we need to determine the vertex form of a parabola and substitute the given information to find the specific equation(s).
The vertex form of a parabola is given by: y = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola.
Given that the vertex of the parabola is on the line y = 3x + 1, we can equate the two equations:
3x + 1 = a(x - h)^2 + k
Since the parabola has the same shape and direction as the graph of y = 3x^2, it means that a = 3.
Substituting a = 3 into the equation, we get:
3x + 1 = 3(x - h)^2 + k
To find the vertex (h, k), we use the information that (1,10) lies on the parabola. Therefore, we substitute x = 1 and y = 10 into the equation:
3(1) + 1 = 3(1 - h)^2 + k
4 = 3(1 - h)^2 + k
Now, we have two equations:
3x + 1 = 3(x - h)^2 + k
4 = 3(1 - h)^2 + k
Let's simplify the equations further:
1) 3x + 1 = 3(x^2 - 2xh + h^2) + k
3x + 1 = 3x^2 - 6xh + 3h^2 + k
2) 4 = 3(1 - h)^2 + k
We can now solve these equations simultaneously to find the values of h, k, and the resulting equation(s) of the parabola.
Solving equation 2) for k:
4 = 3(1 - h)^2 + k
4 = 3(1 - 2h + h^2) + k
4 = 3 - 6h + 3h^2 + k
3h^2 - 6h + (k - 1) = 0
Comparing this equation with the standard form of a quadratic equation (ax^2 + bx + c = 0), we get:
a = 3
b = -6
c = k - 1
Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / (2a), we can find the values of h:
h = (-(-6) ± sqrt((-6)^2 - 4 * 3 * (k - 1))) / (2 * 3)
h = (6 ± sqrt(36 - 12k + 12)) / 6
h = (6 ± sqrt(48 - 12k)) / 6
h = 1 ± sqrt(4 - k) / 2
Now, substituting the value of h into equation 2) and solving for k:
4 = 3(1 - h)^2 + k
4 = 3(1 - (1 ± sqrt(4 - k) / 2))^2 + k
Expanding and simplifying:
4 = 3(1 - 1 ± sqrt(4 - k))^2 + k
4 = 3(± sqrt(4 - k))^2 + k
4 = 3(4 - k) + k
Simplifying further:
4 = 12 - 3k + k
4 = 12 - 2k
2k = 12 - 4
2k = 8
k = 4
Therefore, the value of k is 4. Substituting k = 4 back into the equation for h:
h = 1 ± sqrt(4 - k) / 2
h = 1 ± sqrt(4 - 4) / 2
h = 1 ± 0 / 2
h = 1
So, the vertex of the parabola is (h, k) = (1, 4).
Now, substituting h = 1 and k = 4 into equation 1) to find the resulting equation(s) of the parabola:
3x + 1 = 3(x - h)^2 + k
3x + 1 = 3(x - 1)^2 + 4
3x + 1 = 3(x^2 - 2x + 1) + 4
3x + 1 = 3x^2 - 6x + 3 + 4
3x + 1 = 3x^2 - 6x + 7
Rearranging the equation:
3x^2 - 9x + 6 = 0
So, the equation of the parabola that satisfies the given conditions is 3x^2 - 9x + 6 = 0.