Question 1119588: Do the following for the given f(x).
Identify the vertex and axis of symmetry on the graph of y=f(x)
Graph y=f(x)
Evaluate f(-2) and f(5)
f(x)=x^2-10
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
For a quadratic polynomial function of the form \ =\ ax^2\ +\ bx\ +\ c) , the  -coordinate of the vertex is found by  and the  -coordinate of the vertex is the value of the function at that -value. That is to say that the vertex is located at:
\))
Since
we note that the value of  , the first-degree coefficient, is zero, hence  and we know that the -coordinate of the vertex must also be zero, and then the -coordinate of the vertex must be \ =\ (0)^2\ -\ 10\ =\ -10) . Therefore the vertex is at the point )
For a quadratic polynomial function of the form , the axis of symmetry is described by the equation  where  is the -coordinate of the vertex. Hence, your axis of symmetry is the line:
To wit: the -axis.
If \ =\ x^2\ -\ 10) , then \ =\ a^2\ -\ 10) , \ =\ \(\frac{z^2}{b}\)^2\ -\ 10) . In fact, \ =\ \(\text{the artist formerly known as Prince}\)^2\ -\ 10)
So, to evaluate ) write out \ =\ (x)^2\ -\ 10) in pencil. Then erase every instance of an and replace it with a  . Then do the indicated arithmetic.
John
My calculator said it, I believe it, that settles it
|
|
|
