Find the Parametrization for the left half of the parabola
y=x^2-4x+3
To find the parametrization for the left half of the parabola given by the equation y = x^2 - 4x + 3, we need to restrict the domain of x to only include the values that correspond to the left half of the parabola.
First, let's find the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of x^2 and x, respectively, in the equation y = x^2 - 4x + 3.
In this case, a = 1 and b = -4. So the x-coordinate of the vertex is:
x = -(-4) / (2 * 1) = 4 / 2 = 2
Now substitute this value of x back into the equation to find the y-coordinate of the vertex:
y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1
So the vertex of the parabola is (2, -1).
Since we want the left half of the parabola, we need to consider values of x that are less than 2. Let's parameterize these values of x using a parameter t.
Let t be any real number.
We can set x = 2 - t, which will give us x values ranging from 2 to negative infinity as t increases.
Substituting x = 2 - t into the equation y = x^2 - 4x + 3:
y = (2 - t)^2 - 4(2 - t) + 3
y = 4 - 4t + t^2 - 8 + 4t + 3
y = t^2 - 1
So the parametrization for the left half of the parabola is x = 2 - t and y = t^2 - 1, where t is any real number.