Question 367941: Graph the quadratic function  . State the ordered pairs representing the x-intercept(s), y-intercept (if any), and the vertex of the parabola. If the parabola has a maximum or minimum, give its coordinates.
So far I have the following:
To find the y-intercept set x=0 and solve for f(x)
y-intercept is at (0,5)
To find the x-intercept set x=0 and solve for x
and then I get stuck...is this even right to this point? How do I finish this problem and how do I graph it? Also, how do I know if the parabola has a maximum or minimum and give its coordinates?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Use the quadratic formula where
 ,  , and 
\ \pm\ \sqrt{(-10)^2\ -\ 4(-7)(5)}}{2(-7)})
Roughly
or
For any quadratic function of the form:
\ =\ ax^2\ +\ bx +\ c)
The  -coordinate of the -intercept is
\ =\ a(0)^2\ +\ b(0) +\ c\ =\ c)
hence the graph intercepts the -axis at )
If
is a pair of real numbers (which is true if and only if  ), then the graph intercepts the  -axis at
)
and
)
If
is NOT a pair of real numbers (which is true if and only if  ), then the graph does not intercept the -axis at all.
The -coordinate of the vertex is 
The -coordinate of the vertex is \ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right)\ +\ c)
If the lead coefficient is positive, that is if  , then the parabola opens upward and the vertex is a minimum. If the lead coefficient is negative, that is if  , then the parabola opens downward and the vertex is a maximum.
Finally, the axis of symmetry is the vertical line: 
John
My calculator said it, I believe it, that settles it
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