how do I find the vertex of y = x^2 - 18? I got -9, -81 but that seems impossible I think I'm missing something
y = x^2 - 18.
To get vertex you try to write the equation in the form y=a(x-h)^2 +k, where vertex is (h,k)
Our equation can be written as
y=1(x-0)^2 +(-18)
Comparing with the formula equation, we see that h=0 and k=-18
So vertex (0,-18)
thank you!
To find the vertex of a quadratic equation in the form of y = ax^2 + bx + c, you can use the formula x = -b / (2a) to find the x-coordinate of the vertex. The y-coordinate can then be calculated by substituting the x-value into the equation.
In the given equation, y = x^2 - 18, we can see that a = 1, b = 0, and c = -18.
Using the formula x = -b / (2a), we can substitute the values and solve for x:
x = -0 / (2 * 1)
x = 0 / 2
x = 0
Now, substituting this value of x back into the equation to find the y-coordinate:
y = (0)^2 - 18
y = -18
Therefore, the vertex of the quadratic equation y = x^2 - 18 is (0, -18).
It appears that you made a mistake in your calculation. The correct vertex is (0, -18), not (-9, -81).