SOLUTION: Find the vertex, line of symmetry, the max/min value of the quadratic function and graph the function.
f(x)= -2x^2+2x+8
I have tried to solve this several times and I keep ge
Algebra.Com
Question 347826: Find the vertex, line of symmetry, the max/min value of the quadratic function and graph the function.
f(x)= -2x^2+2x+8
I have tried to solve this several times and I keep getting stuck. This is what I have tried:
putting it into the form ax^2+bx-c to complete the square
-2x^2+2x=-8
Divide both sides by -2 to make x^2 a value of 1
-2x^2+2x/-2 = -8/-2
x^2+x = -4
(1/2)^2 = 2/4
1/4 = 1/2
This is about as far as I can seem to go. I don't understand this stuf at all and I have reviewed many different sites as well as the information from my class and I am just lost. Any help will be very appreciated.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The vertex of a parabola described by the function
\ =\ ax^2\ +\ bx\ +\ c)
has an 
-coordinate of 
and a 
-coordinate of )
. The axis of symmetry is the vertical line 
. If the lead coefficient is positive, the parabola opens upward. Therefore the value of the function at the vertex (i.e. the -coordinate as discussed above) is the minimum value of the function. If the lead coordinate is less than zero, then the parabola opens downward and the vertex is a maximum.
For your function, the -coordinate of the vertex is }\ =\ \frac{1}{2})
. You can calculate )
to get the -coordinate of the vertex. The rest of your questions fall right in line once you have the coordinates of the vertex.
John
My calculator said it, I believe it, that settles it
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