Questions on Geometry: Points, lines, angles, perimeter answered by real tutors!

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Find the vertex, line of symmetry, the maximum or minimum value of the quadratic equation, and graph the function. f(x)= -2x^2+2x+6
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set f(x) = 0 to find where it crosses the x-axis.
-2x^2+2x+6=0
x^2 - x - 3 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation ax^2+bx+c=0

(in our case 1x^2+-1x+-3 = 0

) has the following solutons:

$x[12] = (b+-sqrt( b^2-4ac ))/2\a$

For these solutions to exist, the discriminant b^2-4ac

should not be a negative number.

First, we need to compute the discriminant b^2-4ac

.

Discriminant d=13 is greater than zero. That means that there are two solutions: $x[12] = (--1+-sqrt( 13 ))/2\a$ .

$x[1] = (-(-1)+sqrt( 13 ))/2\1 = 2.30277563773199$
$x[2] = (-(-1)-sqrt( 13 ))/2\1 = -1.30277563773199$

Quadratic expression 1x^2+-1x+-3

can be factored:
1x^2+-1x+-3 = (x-2.30277563773199)*(x--1.30277563773199)

Again, the answer is: 2.30277563773199, -1.30277563773199. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 1*x^2+-1*x+-3 )

The vertex is at the minimum: set the 1st derivative to 0
2x-1=0
x = 1/2
---------
sub for x in the eqn
y = (1/4) - 1/2 - 3
y = -3 1/4 = -13/4
So the vertex is at (1/2,-13/4)
Since there's no xy term, the axis of symmetry is parallel to the y-axis. It goes thru the vertex, so it's:
x = 1/2