I need to find the vertex of the parabola y=9x^2+7 but I am confused on what to use for b. I know a=9 but am unsure if b=7. Does b=7?
since there is no x term, b=0
y = 9x^2 + 0x + 7
9 x^2 + 0 x + 7 = y
Well, first sketch a graph of this parabola.
Note that it is symmetric about the y axis. In other worda it is the same fo x = -537 and x = + 537 or whatever number because x^2 = (-x)^2 and there is no b x term.
so the vertex is where x = 0
then y = 7
so (0,7)
now do it right
9 x^2 = y-7
x^2 = (y-7)/9
(x-0)^2 = (1/9)(y-7)
so
vertex at (0 , 7) as we knew
So if I put 0 in a b I would get 0/18 which equals 0 for x?
Hmmm, I am not sure what 18 has to do with the affair.
I'm using the formula -b/2a for the function9x^2+7. So I would get -0/2(9) which equals -0/18 and the final result is 0.
Oh, ok, anyway zero is the right answer.
That is just fine. I completed the square because I am a creature of habit.
Thanks
To find the vertex of a parabola in the form y = ax^2 + bx + c, you can use the formula:
x = -b / (2a)
In this case, the equation y = 9x^2 + 7 is already in the form of y = ax^2 + bx + c, where a = 9 and b = 0 (since there is no x term).
Therefore, in this equation, b is actually 0, not 7. To find the vertex, substitute the values of a and b into the formula:
x = -0 / (2 * 9) = 0 / 18 = 0
So, the x-coordinate of the vertex is 0.
To find the y-coordinate of the vertex, substitute the value of x into the original equation:
y = 9(0)^2 + 7 = 0 + 7 = 7
Therefore, the vertex of the parabola y = 9x^2 + 7 is (0, 7).