Work out the Perimeter of the Triangle For 100 Points
1 answer:
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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)
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)