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The graph of [tex]y = ax^2 + bx + c[/tex] is a parabola. The axis of symmetry is [tex]x = -b/2a[/tex]. What are the coordinates of the vertex?

Sagot :

facundo3141592

The vertex can be written as:

(-b/2a, b^2/(4*a) - b^2/2a + c)

For a general parabola:

y = a*x^2 + b*x + c

We can write the vertex as:

(h, k)

The x-value of the vertex is the value of the axis of symmetry.

Then we have:

h = x = -b/2a

Now we need to find the y-value of the vertex.

To do that, we just replace the variable "x" by the x-value of the vertex in our equation, so we get:

k = y = a*(-b/2a)^2 + b*(-b/2a) + c

k = b^2/(4*a) - b^2/2a + c

Then the coordinates of the vertex are:

(h, k) = (-b/2a, b^2/(4*a) - b^2/2a + c)

If you want to read more:

https://brainly.com/question/24302770

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