SOLUTION: find the vertex, the line of symetry, the max/min value of the quadratic unction and graph. f(x)=2x^2-8x+5 vertex= line of symetry equation= max or min of f(x)=2x^2-8x+5 is it

Question 515325: find the vertex, the line of symetry, the max/min value of the quadratic unction and graph.
f(x)=2x^2-8x+5
vertex=
line of symetry equation=
max or min of f(x)=2x^2-8x+5
is it max/min=
graph=
thank you!
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
f(x) = 2x^2 -8x +5
.
The vertex has the x coordinate = -b/2a.
Solve for 'y' to get the ordered pair.
.
f(x) = y
.
x = -b/2a = -(-8)/(2*2) = 8/4 = 2
.
y = 2(2^2) -8*2 +5
y = 8 -16 +5
y = -3
.
The vertex is (2,-3).
.
We know the parabola opens 'up' so the vertex will be the minimum value of f(x).
.
The line of symmetry is the vertical line that passes through the vertex, dividing the parabola into two symmetric halves.
.
x = 2 would do it. Note that for any value of 'y', x =2. The slope is undefined because x-x₁=0, by definition, so the line is vertical.
.
To determine the maximum of minimum, you also can take the first deriviate and set it = 0.
.
dy/dx = 4x-8
4x = 8
x = 2
.
The maximum (or minimum) occurs at x=2. We know from above that the vertex is (2,-3). So that confirms it is a minimum.
.

.
You also should consider the roots.
y = 2x^2-8x+5 cannot be factored.
So you can use the quadratic equation.
.
(-b +sqrt(b^2-4ac)) / 2a and (-b -sqrt(b^2-4ac)) / 2a
.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=24 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3.22474487139159, 0.775255128608411. Here's your graph:

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