Like graphing quadratic equations

## I need help solving something

# 1 Answer

y = x^{2} - 6x + 4

Use an equation that will help us find the x of our vertex point (x,y)

equation for x of your vertex = -b/2a

this equation

a = 1, b = -6, c = 4

vertex x = -(-6)/2(1) = 6/2 = 3

Take the x you just found and plug it back into original equation to find the y of your vertex

y = (3)^{2} - 6(3) + 4 = 9 - 18 + 4 = -5

Vertex is: (3,-5)

Now you can chose one or two points on either side of your vertex x and plug in to find a couple more points. Because vertex x = 3, lets use 2 and 4

y = (2)^{2} - 6(2) + 4 = 4 - 12 + 4 = -4 Point = (2,-4)

y = (4)^{2} - 6(4) + 4 = 16 - 24 + 4 = -4 Point = (2,-4)

If you need a couple more points then choose x to be 1 and 5.

# Comments

I have the vertex down just I need to see how to graph them like I know how to graph the vertex just graphing from the axis symmetry?

Well the axis of symmetry will be straight up and down from your vertex any time you have a y = x^{2} type of equation.

You can either choose points like I did above on either side of your x of your vertex and create a table, then graph off of that table.

x: 1 2 3 4 5

y: -1 -4 -5 -4 -1

OR the general rule for graphing parabolas (starting at your vertex)

1) Right 1 and 1a up or down (up if a>0, down if a<0)

and Left 1 and 1a up or down

For your equation:

vertex: (3, -5)

Right 1 and 1(1) [a for this equation is 1], point: (4,-4)

Left 1 and 1(1), point: (2,-4)

2) Right 2 and 4a up or down (up if a>0, down if a<0)

and Left 2 and 4a up or down

For your equation:

vertex: (3, -5)

Right 2 and 4(1) [a for this equation is 1], point: (5,-1)

Left 2 and 4(1), point: (1,-1)

## Comments

Can you please give me a specific problem?

y=x^2-6x+4

Comment