how to graph a parabola of y=(x-1)(x-5)?

Question

i neeeeeeeed help

Answer 1

Given y = (x-1)(x-5)

y = x^2-5x-x+5

y = x^2-6x+5

Compare it to standard parabola equation y = ax^2+bx+c

Graph of above parabola y = x^2-6x+5

Answer 2

y  = (x  - 1)(x  - 5)

y  = x ² - 6x  + 5

The above equation represent a standard form of parabola y  = ax ² + bx + c .

a  = 1, b  = -6, c  = 5

Axis of symmetry x  = -b /2a  = 6/2

Axis of symmetry x  = 3

Substitute x  value in y  = x ² - 6x  + 5

y  = (3)² - 6(3) + 5

y  = -4

Vertex of parabola (3, -4)

Choose values for x and find the corresponding values for y.

x	y  = x 2 - 6x  + 5	(x, y)
1	y = (1)2 - 6(1) + 5 y = 0	(1,0)
2	y = (2)2 - 6(2) + 5 y = -3	(2,-3)
4	y = (4)2 - 6(4) + 5   y = -3	(4,-3)
5	y = (5)2 - 6(5) + 5 y = 0	(5,0)

1.Draw a coordinate plane.

2.Plot the axis of symmetry, vertex and coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.