SOLUTION: Find the domain and range of the quadratic equation f(x)=x^2-8x-9

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Question 991271: Find the domain and range of the quadratic equation f(x)=x^2-8x-9
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

The domain of any polynomial function is the set of real numbers.

The lead coefficient of your quadratic is positive, therefore the graph is a parabola that opens upward. Hence the minimum value is the value of the function at the vertex. Find the -coordinate of the vertex by dividing the opposite of the first-degree term coefficient by two times the lead coefficient and then finding the value of the function by evaluating the function at the -coordinate of the vertex. The range is then all real numbers greater than or equal to the value of the function at the vertex.

John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website!
.
Hello,
your question is incorrect.

The correct question is:

        Find the domain and range of the quadratic function   f(x)=x^2-8x-9.

An equation has no domain. The term "domain" is not defined for equations.
It is defined for functions.
It is not applicable for equations. It is applicable for functions.

The same is for range. It is for functions, not for equations.

OK. Now I am ready to answer this question:

        Find the domain and range of the quadratic function f(x)=x^2-8x-9.

1) The domain of a quadratic function is the entire number line, i.e the set of all real numbers.
      It is true for any quadratic function. Particularly, it is true for the given function.

2) The range of a quadratic function is stretched from its minimal value to the positive infinity, if the parabola is U-shaped (has positive coefficient at ,
      as it is in your case).
      If the parabola is bottom-up (the quadratic function has negative coefficient at ), then its range is stretched from its maximal value to the negative infinity.

      So, in your case we need to find the minimal value of the quadratic function. For a quadratic function f(x) = the minimum is reached at x = -
      and is equal to f(-). In your case   - = = 4 and f(4) = = 16 - 32 - 9 = -25.

      Therefore, the range of your function is semi-infinite segment   [-25, ).