I am familiar with finding the vertex of a polynomial but I don't know how to solve this equation.
Give the vertex of x^2+9
can you please explain in steps?
Thank you.
An equation requires an equal sign.
If you mean
y = x^2+9
you can see that since x^2 can never be less than zero, the vertex is at x=0. Since y=9 there, the vertex is at (0,9)
Or, recalling that the vertex of
y = (x-h)^2 + k
is at (h,k), note that we have h=0 and k=9.
To find the vertex of the equation x^2 + 9, you can follow these steps:
1. Identify the coefficients: In this equation, the coefficient of the x^2 term is 1, and the constant term is 9.
2. Recall that the vertex of a quadratic equation in the form y = ax^2 + bx + c is given by the formula:
x = -b / (2a)
y = f(x), where f(x) is the value of the equation at x.
3. Plug in the values: For the equation x^2 + 9, a = 1, b = 0 (as there is no x term), and c = 9.
4. Calculate x: Using the formula x = -b / (2a), substitute the respective values:
x = -(0) / (2 * 1) = 0
5. Calculate y: Plug the calculated value of x back into the equation to find y:
y = (0)^2 + 9 = 9
Thus, the vertex of the equation x^2 + 9 is (0, 9).