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Use the quadratic equation to get the roots

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation ax^2+bx+c=0

(in our case 1x^2+-5x+4 = 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=9 is greater than zero. That means that there are two solutions: $x[12] = (--5+-sqrt( 9 ))/2\a$ .

$x[1] = (-(-5)+sqrt( 9 ))/2\1 = 4$
$x[2] = (-(-5)-sqrt( 9 ))/2\1 = 1$

Quadratic expression 1x^2+-5x+4

can be factored:
1x+-5x+4 = 1(x-4)*(x-1)

Again, the answer is: 4, 1. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 1*x^2+-5*x+4 )

In order to find the vertex point you need to complete the square. You take the middle term half it and then square it.
y=x^2 - 5x + 4
Subtract 4 from each side
y-4=x^2 - 5x take -5 and divide by 2 -5/2 and then square this term 25/4
y-4+25/4=(x-5/2)^2
y-16/4+25/4=(x-5/2)^2
y+9/4=(x-5/2)^2
Subtract 9/4 from both sides
y=(x-5/2)^2 -9/4
The vertex o=is (5/2,-9/4)
We know that since a is positive the parabola is facing upward
the directrix is below the vertex at distance -c on the x axis
the focus is a distance +c above the vertex
a=coefficient on x^2 term
It is 1 The equation for determining the c value is a=1=1/4c

4c=1
c=1/4
Y value for vertex+1/4= Focus, x value for vertex
the directrix is a line y=8/4=2