Ineed another graphing quadratic pointer f(x)=x square -18x +80

F(x)=x^2-18x+80
ok so remember
you need to know
1. degree of function- tells what kind of graph
2. leading term- tells which direction graph opens
3. xintercept and y intercepts
4. vertex

ok so in form
f(x)=ax^2+bx+c 

1.
degree, 2nd degree so parabola

2.
a=leading terma=+1 in this case so the graph opens up and is a normal graph

3.xintercept is where graph crosses x axis or where y=0, y=f(x) so
0=x^2-18x+80
0=(x-8)(x-10)
xints are (8,0) and (10,0)
y intercept is where cross y axis or where x=0
set x=0
yint=0^2-18(0)+80
yint=80
yint=(0,80)

4. vertex (vertex is highest point if graph opens down, and lowest point if graph opens up)
if in form f(x)=ax^2+bx+c, the x value of the vertex is given by -b/(2a) so
b=-18
a=1
-(-18)/2(1)
18/2=9
input
9^2-18(9)+80=yvalue
-1=yvalue
vertex=(9,-1)

so we have
1. graph is parabola
2. opens up
3. crosses through points (8,0), (10,0), (0,80)
4. vertex=(9,-1)

so we plot the points (8,0), (10,0), (0,80) and (9,-1)  and draw a prabola opening up with (9,-1) as the highest point
attachment below ( it might be hard to graph (0,80) since it is so high up so you don't have to graph that point)

Idon’t know actually is more difficult