-8x^2-3x+5+2x^2+7x-5
1 answer:
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Answers:
1) Graph Line 1
a) Graph: Please, see the attached graph.
b) (Line 1) Main Street (Equation in slope intercept form): y=(1/3)x+4
2) Graph Line 2
a) Graph: Please, see the attached graph.
b) Y intercept: (0,10)
c) X intercept: (5,0)
d) (Line 2) Street Equation in Standard Form: y+2x=10
3) Grap Line 3
a) Graph: Please, see the attached graph.
b) (Line 3) Street Equation in Point Slope Form: <u>y-3=(5/3)(x+3)</u>
Street Equation in Slope Intercept Form: <u>y=(5/3)x+8</u>
Solution:
1) Graph Line 1: (Main Street)
a) Graph the line formed from the two given points (9,7) (-9,1)
Please, see the attached graph.
b) Find the street equation in slope intercept form
P1=(9,7)=(x1,y1)→x1=9, y1=7
P2=(-9,1)=(x2,y2)→x2=-9,y2=1
m=(y2-y1)/(x2-x1)
m=(1-7)/(-9-9)
m=(-6)/(-18)
m=6/18
m=(6/6)/(18/6)
m=1/3
Point Slope form:
y-y1=m(x-x1)
y-7=(1/3)(x-9)
Slope Intercept Form:
y-7=(1/3)x-(1/3)(9)
y-7=(1/3)x-(1)(9)/3
y-7=(1/3)x-9/3
y-7=(1/3)x-3
y-7+7=(1/3)x-3+7
y=(1/3)x+4
2) Graph Line 2
a) Graph the line given the x and y intercepts
y intercept: 10
x intercept: 5
Please, see the attached graph
b) Y intercept
y intercept: b=10
Y intercept: (0,b)→Y intercept: (0,10)
c) X intercept
x intercept: a=5
X intercept: (a,0)→X intercept: (5,0)
d) Find the Street Equation in Standard Form
P1=(0,10)=(x1,y1)→x1=0, y1=10
P2=(5,0)=(x2,y2)→x2=5, y2=0
m=(y2-y1)/(x2-x1)
m=(0-10)/(5-0)
m=(-10)/(5)
m=-2
Point Slope Form:
y-y1=m(x-x1)
y-10=(-2)(x-0)
y-10=(-2)(x)
y-10=-2x
Standard Form:
y-10+10+2x=-2x+10+2x
y+2x=10
3) Graph Line 3
a) Graph the line given a point and the slope. (-3,3); m=5/3
Point 1: P1=(-3,3)=(x1,y1)→x1=-3, y1=3
Slope: m=5/3
m=dy/dx
5/3=dy/dx→dy=5, dx=3
Point 2: P2=(x2,y2)
x2=x1+dx→x2=-3+3→x2=0
y2=y1+dy→y2=3+5→y2=8
Point 2: P2=(x2,y2)→P2=(0,8)
Please, see the attached graph.
b) Find the Street Equation in Point Slope Form
Point Slope Form:
y-y1=m(x-x1)
y-3=(5/3)(x-(-3))
y-3=(5/3)(x+3)
Slope Intercept Form:
y-3=(5/3)x+(5/3)(3)
y-3=(5/3)x+(5)(3)/3
y-3=(5/3)x+15/3
y-3=(5/3)x+5
y-3+3=(5/3)x+5+3
y=(5/3)x+8
