Vertex Form of a Quadratic Equation
Key Questions

Answer:
#" "#
Please read the explanation.Explanation:
#" "#
Quadratic Equations in Vertex Form have a general form:#color(red)(y=f(x)=a(xh)^2+k# , where#color(red)((h,k)# is the#color(blue)("Vertex"# Let us consider a quadratic equation in Vertex Form:
#color(blue)(y=f(x)=(x3)^2+8# , where#color(green)(a=1; h=3; k=8# Hence,
#color(blue)("Vertex "= (3, 8)# To find the yintercept, set
#color(red)(x=0# #y=(03)^2+8# #y=9+8# #y= 17# Hence, the yintercept:
#color(blue)((0, 17)# We can use a table of values to draw the graph:
Use the table with two columns
#color(red)(x and y# to draw the graph as shown below:The Parent Graph of
#color(red)(y=x^2# can also be seen for comparison, to better understand transformation.Also note that,
Axis of Symmetry is
#color(red)(x=h# #rArr x= 3# We can verify this from the graph below:
Hope it helps.

Since the equation is:
#y=x^2+bx+c# the vertex is
#V(b/(2a),Delta/(4a))# ,or, found the
#x_v=b/(2a)# you can substitue it in the equation of the parabola at the place of#x# , finding the#y_v# . 
Vertex Form
#y=a(xh)^2+k# ,where
#(h,k)# is the vertex.
I hope that this was helpful.