Chapter 1 Test (7-10)

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1 Test (7-10)

Chapter 1 has been completed. For those who want to take the Chapter 1 test to get extra practice, see attachment. Enjoy.

[Eigenvalue]

7. We can use slope-intercept form to solve this problem
y-y1=m(x-x1)
y+4=-2(x-3)
y=-2x+2
y+2x=2

8. The equation of a circle is (x-h)²+(y-k)²=r²
(x-4)²+(y+3)²=25
x²-8x+16+y²+6y+9=25
x²-8x+y²+6y=0

9. We can complete the square to find the center and radius
x²+4x+y²-2y=4
x²+4x+4+y²-2y+1=9
(x+2)²+(y-1)²=9
The center is (-2,1) and the radius is 3

10. 3y=6-2x
y=([tex]\frac{-2}{3}[/tex])x+2
The slope of the line is [tex]\frac{-2}{3}[/tex]
A line that is parallel has the same slope
y+1=([tex]\frac{-2}{3}[/tex])(x-1)
y=([tex]\frac{-2}{3}[/tex])x-[tex]\frac{1}{3}[/tex]
2x+3y=-1
A line that is perpendicular has an opposite reciprocal slope
y-3=([tex]\frac{3}{2}[/tex])x
y=([tex]\frac{3}{2}[/tex])x+3
2y-3x=6

[nycmath]

7. We can use slope-intercept form to solve this problem
y-y1=m(x-x1)
y+4=-2(x-3)
y=-2x+2
y+2x=2

8. The equation of a circle is (x-h)²+(y-k)²=r²
(x-4)²+(y+3)²=25
x²-8x+16+y²+6y+9=25
x²-8x+y²+6y=0

9. We can complete the square to find the center and radius
x²+4x+y²-2y=4
x²+4x+4+y²-2y+1=9
(x+2)²+(y-1)²=9
The center is (-2,1) and the radius is 3

10. 3y=6-2x
y=([tex]\frac{-2}{3}[/tex])x+2
The slope of the line is [tex]\frac{-2}{3}[/tex]
A line that is parallel has the same slope
y+1=([tex]\frac{-2}{3}[/tex])(x-1)
y=([tex]\frac{-2}{3}[/tex])x-[tex]\frac{1}{3}[/tex]
2x+3y=-1
A line that is perpendicular has an opposite reciprocal slope
y-3=([tex]\frac{3}{2}[/tex])x
y=([tex]\frac{3}{2}[/tex])x+3
2y-3x=6

You outdid yourself with yet another detailed answer. Keep up the good work.