vertex of y=x^2-8x+22?

Answer 1

You are given the quadratic function
y = x² - 8x + 22

To find the vertex, change the given equation into the form
y = a(x - h)² + k

Where
a = is a constant
(h,k) is the vertex

Thus,
y = x² - 8x + 22

We complete the square, by squaring the half of the coefficient of x (which is -8, the half of -8 is -4, and the square of -4 is 16.) Thus,
y = x² - 8x + 16 + 6

Now, we can factor
y = (x - 4)² + 6

Therefore, the vertex is
(4,6)

^_^

Answer 2

To calculate the vertex, you have to complete the square:

y = x^2 - 8x + 22

That means you have to take the coefficient of x (which is -8), take half of that, and then square it. So half of -8 is -4, squared is +16. You want to add 16 and subtract 16 (which is, effectively, adding zero and not changing the equation). So now we have

y = x^2 - 8x + 16 + 22 - 16

Now that we have added the "half-squared" of the coefficient of x, the first three terms of the polynomial should be a perfect square. It becomes:

y = (x - 4)^2 + 22 - 16
or
y = (x - 4)^2 + 6

To get the coordinates of the vertex, you take the negative of the term in the brackets, and the whole number outside of the brackets. In this case, it's (- (-2),6), or (2,6).

So the vertex is (2,6)

Answer 3

the vertex of the equation will be the point at which the parabola has a maximum or minimum depending of the concavity. to find this point take the first derivative and set it to zero. 2x -8 = 0. x= 4. plug this into the original eqn and you get y= 6. vertex = (4, 6)

Answer 4

y' = 2x-8 = 0

2x = 8

x = 4

y(4) = 16-32+22 = 6

Vertex: (4, 6)

Answer 5

y' = 2x-8 = 0
2x = 8
x = 4
y(4) = 16-32+22 = 6
Vertex: (4, 6)

Answer 6

it has lots of answers.for each x ,you have an answer to y.

Answer 7

(4,6)

Answer 8

(4,6)