How do you use the important points to sketch the graph of y = x^2 − 6x + 1?

1 Answer
Feb 8, 2018

color(blue)("Y-int": (0,1))
color(red)("Vertex": (3,8))
color(green)("X-Intercept/Zero":(0.172,0) and (5.828,0)

Explanation:

The most important points should be the color(red)"vertex", the color(blue)("y-int"), and the color(green)"zeroes" (if there are any).

The x-coordinate of the color(red)(vertex) of any quadratic equation
y=ax^2+bx+c is:

(-b)/(2a)

Plug in the x-coordinate back into the equation to find y. Do it on this equation:

b=-6

a=1

c=1

(-(-6))/(2a)=3

Now you have the color(red)(vertex) as (3,y)

y=(3^2)-6*(3)+1=-8

The color(red)(vertex) is (3, -8)

The color(blue)("y-int") of the quadratic equation
y=ax^2+bx+color(orange)(c) is simply color(orange)(c).

The color(blue)("y-int") of this equation is (0,1).

To find the color(green)"zeroes", plug into the quadratic formula, which is given by:

(-b+-sqrt(b^2-4ac))/(2a)

Plug in:

(6+-sqrt((-6)^2-4*1*1))/(2*1)

Simplify:

(6+-4sqrt(2))/2

3+-2sqrt(2)

The color(green)"zeroes" are: (5.828,0) and (0.172,0)