Identify the vertex of y=4(x−25)2−61 .
Show your work
The equation is in the form y = a(x-h)^2 + k, where (h,k) represents the vertex of the parabola.
Comparing the given equation, y = 4(x-25)^2 - 61, with this form, we can see that the vertex is at (h,k) = (25, -61).
Therefore, the vertex of y = 4(x-25)^2 - 61 is (25, -61).
Identify the vertex of y=4(x−25)^2−61 .
Show your work.
To identify the vertex of the equation y = 4(x-25)^2 - 61, we can directly compare it with the general vertex form of a parabola: y = a(x-h)^2 + k, where (h,k) represents the vertex.
In this case, we can see that the parabola is already in the vertex form with h = 25 and k = -61.
Therefore, the vertex of y = 4(x-25)^2 - 61 is (25, -61).
Identify the vertex of y=4(x−25)2−61 .
Explain your work
To identify the vertex of the equation y = 4(x-25)^2 - 61, we can use the vertex formula for parabolas of the form y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
In the given equation, we have (h, k) = (25, -61) as the vertex.
The equation tells us that the vertex is located at the point (25, -61), so the vertex of y = 4(x-25)^2 - 61 is (25, -61).
Identify the vertex of y=4(x−25)2−61 .
Show me how you got it
To identify the vertex of the equation y = 4(x-25)^2 - 61, we can use the vertex formula for parabolas of the form y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, the equation is already in the form y = 4(x-25)^2 - 61, so we can directly read the values of h and k.
Comparing the given equation with the vertex form, we have h = 25 and k = -61.
Therefore, the vertex of y = 4(x-25)^2 - 61 is (25, -61).
Determine the zero(s) of the quadratic function y=6x2+13x+5 . Show all steps.
To determine the zeros of the quadratic function y = 6x^2 + 13x + 5, we need to find the values of x for which y is equal to zero.
Step 1: Set y = 0
0 = 6x^2 + 13x + 5
Step 2: Factor or use the quadratic formula to solve for x.
Since the equation does not easily factor, we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 6, b = 13, and c = 5.
x = (-13 ± √(13^2 - 4*6*5)) / (2*6)
x = (-13 ± √(169 - 120)) / 12
x = (-13 ± √49) / 12
x = (-13 ± 7) / 12
So we have two possible solutions:
x = (-13 + 7) / 12 = -6/12 = -1/2
x = (-13 - 7) / 12 = -20/12 = -5/3
Therefore, the zeros of the quadratic function y = 6x^2 + 13x + 5 are x = -1/2 and x = -5/3.
To identify the vertex of the quadratic function y = 4(x - 25)^2 - 61, we can use the vertex form of a quadratic equation, which is y = a(x - h)^2 + k. In this equation, (h, k) represents the coordinates of the vertex.
Comparing the given equation y = 4(x - 25)^2 - 61 with the vertex form, we see that a = 4, h = 25, and k = -61.
Therefore, the vertex coordinates are (h, k) = (25, -61).
So, the vertex of the quadratic function y = 4(x - 25)^2 - 61 is (25, -61).
Note: In general, to find the vertex of a quadratic equation in standard form (y = ax^2 + bx + c), we can use the formula x = -b/2a to find the x-coordinate of the vertex, and then substitute this value into the equation to find the corresponding y-coordinate.