SOLUTION: find the vertex, and the line of symmetry, the maximum or minimum value of the quadratic function. then graph.
f(x)-2x^2+2x+8
x-vertex=
y-vetex=
line of symmetry=
max/min
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Question 398542: find the vertex, and the line of symmetry, the maximum or minimum value of the quadratic function. then graph.
f(x)-2x^2+2x+8
x-vertex=
y-vetex=
line of symmetry=
max/min value=
is the value min or max=
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
find the vertex, and the line of symmetry, the maximum or minimum value of the quadratic function. then graph.
f(x)-2x^2+2x+8
x-vertex=
y-vetex=
line of symmetry=
max/min value=
is the value min or max=
..
Before I begin, I will assume a typing error in the given problem. After f(x) there should be an equal(=) sign instead of a minus (-) sign.
..
The first step is to write equation to standard form: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. An alternative method is to use the formula, -b/2a, to find the x-coordinate, then plugging it back into the equation to find the y-coordinate of the vertex. a is the coefficient of x^2, and b, is the coefficient of x. I will be doing the former method.
..
f(x)=2x^2+2x+8
complete the square
y=2(x^2+x+1/4)+8-1/2
y=2(x+1/2)^2+15/2
This is shown to be a parabola opening upward with vertex at (-1/2,15/2)
The line of symmetry is a vertical line,x=-1/2. It has a minimum value=15/2
In summary:
vertex(-1/2,15/2)
line of symmetry x=-1/2
minimum value 15/2
The following is a graph of the parabola
..
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