Graph the function?

Question

f(x)=x(sq)-2x+5

Answer 1

x^2-2x+5
the table will be
x= -15 -10 -5 0 5 10 15
y= 260 125 40 5 20 85 200
plot now taking a suitable scale

if you want you can have smaller values

x= -3 -2 -1 0 1 2 3
y= 20 13 8 5 4 5 8

Answer 2

f(x)=x(sq)-2x+5
The roots of this equation are
[2 +sqrt(4-4*1*5)]/2 = [2+sqrt(-16)]/2 = 1+ 2i
The other root must be 1-2i. The equation is of the second degree and so is a parabola. Because the sign of x^2 is positive, the parabola is concave down (like a smile). Because the roots are imaginary, the curve lies entirely above the y-axis
Its vertex is at (1,4) and so its minimum value is 4. The axis of symmetry is x=1. (Axis of symmetry is always at -(b/2a) where b is the coefficient of x and a is the coefficient of x^2)

Answer 3

Chose any number for x and complete the function. If you chose 1, you would square 1, subtract 2(1) and add 5. That is 4, which would be your y-coordinate. Choose a couple more numbers for x and complete the function for each. Graph these x,y points on your cartesian plane and connect them with a solid line.

Answer 4

complete the square...so...
f(x) = (x^2 - 2x + 1) + 5 - 1
= (x - 1)^2 +4
---this is now in "y = (x - h)^2 + k" form where (h,k) is the vertex
so...
(1,4) is the vertex
---since the "x^2" term is positive, it opens up
axis of symmetry is vertical and it crosses through the vertex so the axis of symmetry is x = 1.
---plot more points to make the graph more accurate
(or just use a graphing calculator)

Answer 5

It looks like x^2 shifted right one unit and up four units. You can turn the equation into f(x)=(x^2-2x+1)^2+4=(x-1)^2+4.

Answer 6

it is always going to be straight line. U can find the angle of it to x axis by a simple formula