SOLUTION: 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
Algebra.Com
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
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