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
To find the equation(s) of all possible parabolas, you need to use the given information about the vertex and the fact that the parabola has the same size, shape, and direction as the graph of y = 3x^2.
Let's start by finding the coordinates of the vertex of the parabola. Since the vertex lies on the line y = 3x + 1, it will have the same x-coordinate as any point on the line. We can equate the x-coordinate of the vertex with the x-coordinate of the given point (1, 10):
x-coordinate of vertex = 1
Now, to find the y-coordinate of the vertex, substitute the x-coordinate of the vertex into the equation of the line:
y = 3x + 1
y = 3(1) + 1
y = 4
So, the coordinates of the vertex are (1, 4).
Next, we know that the parabola has the same size, shape, and direction as the graph of y = 3x^2. This means that the coefficient of x^2 in the equation of the parabola will be the same, which is 3.
Now, we can use the vertex form of a parabola equation: y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
Plugging in the values, we get:
y = 3(x - 1)^2 + 4
Expanding the equation, we have:
y = 3(x^2 - 2x + 1) + 4
y = 3x^2 - 6x + 3 + 4
y = 3x^2 - 6x + 7
Therefore, the equation of the parabola that satisfies the given conditions is y = 3x^2 - 6x + 7.
Note: There might be another possible equation if the parabola is reflected across the line y = 3x + 1. But based on the given information, we assume the parabola is not reflected.