Question

Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the​ parabola's axis of symmetry. Use the graph to determine the domain and range of the function.
f(x)=6x−x2−25

Answer 1

The vertex is at (3, -16), so we conclude that:

• domain: set of all real numbers.
• range: f(x) ≤ -16.

How to get the vertex of the quadratic function?

For a general quadratic function:

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

The vertex is at:

$x = -b/2a$

In our case, we have:

$f(x) = -x^2 + 6x - 25$

So the x-value of the vertex is:

$x = -6/2*(-1) = 3$

And the y-value of the vertex is:

$f(3) = -(3)^2 + 6*3 - 25 = -9 + 18 - 25 = -16$

So the vertex is at (3, -16)

Also, notice that the function has a negative leading coefficient, which means that the vertex is a maximum.

From that, we conclude that the range is:

f(x) ≤ -16.

And the domain like in all quadratic functions is the set of all real numbers.

The graph can be seen below.

If you want to learn more about quadratic functions:

https://brainly.com/question/1214333

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