Identify the vertex of y=4(x−25)^2−61
The given equation is in vertex form, which is expressed as y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
By comparing the given equation y = 4(x-25)^2 - 61 with the general vertex form, we can deduce that the vertex is located at the point (25, -61).
Therefore, the vertex of y = 4(x-25)^2 - 61 is (25, -61).
To identify the vertex of the given quadratic equation, we can make use of the vertex form, which is written as:
y = a(x-h)^2 + k
In this form, the vertex of the parabola is located at the point (h, k).
Now let's compare the given equation, y = 4(x−25)^2−61, with the vertex form. We can see that h = 25 since the expression being squared is (x - 25). To determine k, we can simply look at the constant term, which is -61.
Therefore, the vertex is located at the point (25, -61).
Hence, the vertex of the quadratic equation y = 4(x−25)^2−61 is (25, -61).
To identify the vertex of the quadratic equation y = 4(x - 25)^2 - 61, we need to determine the values of x and y that make the equation true.
The given equation is in the form y = a(x - h)^2 + k, where (h, k) represents the vertex of the quadratic function.
Comparing the equation to the standard form, we find that h = 25, which indicates that the x-coordinate of the vertex is 25. To find the y-coordinate, we substitute x = 25 into the equation:
y = 4(25 - 25)^2 - 61
= 4(0)^2 - 61
= 4(0) - 61
= 0 - 61
= -61
Hence, the vertex of the quadratic equation y = 4(x - 25)^2 - 61 is (25, -61).