Graphing Quadratic Functions
Key Questions

When you graph a quadratic, there are a couple of things you need to consider that will make your life easier.
Firstly, you need to find the vertex . This is derived through (wait for it) vertex form. In the equation
#y=a(xh)^2+k# , the point#(h,k)# is your vertex. It is either the highest or lowest point on your graph, and it is the first thing you need to graph.For example, consider the equation
#y = 2(x6)^2+2# Here
#h = 6# and#k = 2# , therefore the vertex would be#(6,2)# Secondly, you need to find the direction of the graph. If your
#a# value in the above equation is negative , then the equation opens downwards. If it is positive, then it opens upwards. This will help you get a good picture of what your graph will look like.Considering our sample equation,
#a = 2# , therefore it opens down.Thirdly, you need to find the intercepts of the graph. If you plug in 0 for
#y# in the equation above, and solve for#x# , then you will fnd the coordinates of the two intercepts. These will give you additional points to complete your curve.Now let's find the zeroes of our sample equation:
#0 = 2(x6)^2+2# #2 = 2(x6)^2# #1 = (x6)^2# #+1 = x6# #x = 6+1 = {7, 5}# Hence our intercepts would be the points
#(7,0)# and#(5,0)# Lastly, you just need to draw a curve through all these three points, and you are done! This is what it would look like:
If you are required to find more coordinates, the easiest way to get them is to just plug the equation into your calculator, and get more points. If you need to do this by hand, then you'd just plug in more
#y# 's and solve for#x# .
Hope that helped :)

Answer:
Use the formula
#b/(2a)# for the x coordinate and then plug it in to find the y.Explanation:
A quadratic equation is written as
#ax^2+bx+c# in its standard form. And the vertex can be found by using the formula#b/(2a)# .For example, let's suppose our problem is to find out vertex (x,y) of the quadratic equation
#x^2+2x3# .1) Assess your a, b and c values. In this example, a=1, b=2 and c=3
2) Plug in your values into the formula
#b/(2a)# . For this example, you'll get#2/(2*1)# which can be simplified to 1.3) You just found the x coordinate of your vertex! Now plug in 1 for x in the equation to find out the ycoordinate.
4)
#(1)^2+2(1)3=y# .5) After simplifying the above equation you get : 123 which is equal to 4.
6) Your final answer is (1 ,4)!
Hope that helped.

My favorite way is to complete the square in the equation for y, find the vertex and yintercept, and draw the parabola.
Our goal is to make the equation#y=ax^2+bx+c# look like#y=a(xh)^2+k# , then the parabola's vertex (tip) is at#(h,k).# Example:
Say your equation is given as:#y=x^26x+8.# Here#a=1# .
Notice that#(xh)^2# has a middle term of#2hx# ,
so we need to make#h = 6/3 = 2.# So
#(x3)^2 = x^2  6x + 9,# but we want a constant term of 8, meaning#y=(x3)^21.# We read the vertex at
#(3,1)# and the yintercept at#(0,8)# .
This gives a parabola sitting on the vertex and curving upward, with axis of symmetry on the vertical line#x=3.# By the way this example also has roots (xintercepts) at
#(2,0)# and#(4,0).# (How do these relate to the factors of#x^26x+8# ?)You're welcome, and happy graphing from
#dansmath.#