Chapter 1 Test

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Precalculus
Michael Sullivan
Edition 10
Chapter 1 Test

Chapter 1 test is also our review content.

[Eigenvalue]

1. Use the distance formula to find the distance
d=[tex]\sqrt{(3-(-1))²+(-1-5)²}[/tex]

=[tex]\sqrt{4²+6²}[/tex]

=[tex]\sqrt{52}[/tex]

=2[tex]\sqrt{13}[/tex]

[Eigenvalue]

2. Use the midpoint formula
M=([tex]\frac{-1+5}{2}[/tex], [tex]\frac{3+(-1)}{2}[/tex])=(2,1)

[Eigenvalue]

3. Use point slope form to find the equation:
Find the slope: [tex]\frac{-1-3}{5-(-1)}[/tex]=[tex]\frac{-2}{3}[/tex]

Substitute into point slope form:
y-3=[tex]\frac{-2}{3}[/tex](x+1)

y=[tex]\frac{-2}{3}[/tex]x+[tex]\frac{7}{3}[/tex]

Check:
Let x=5 and y=-1
-1=[tex]\frac{-2}{3}[/tex](5)+[tex]\frac{7}{3}[/tex]

-1=[tex]\frac{-3}{3}[/tex]

4. For every 3 unit increase in y-values, the x value decreases by 2 units

[Eigenvalue]

8. Use the formula for the area of a circle: (x-h)²+(y-k)²=r²
(h,k)=(4,-3)
r=5

(x-4)²+(y+3)²=25
x²-8x+16+y²+6y+9=25
x²-8x+y²+6y=0

[Eigenvalue]

5. Point (3,-4); slope=-2

Use point slope form to write the equation:
y+4=-2(x-3)
y+4=-2x+6
y=-2x+2

5[tex]\frac{1}{2}[/tex]. Domain: all real numbers (there are no values of x that would result in division by 0 or a negative value in a radicand)
Range: all real numbers (there are no values of x that would result in division by 0 or a negative value in a radicand)

X-intercepts: 0=-2x+2
2x=2
x=1
(1,0)

Y-intercepts:y=-2(0)+2
y=2
(0,2)

Find the inverse function by switching the x and y values
x=-2y+2
x-2=-2y
y=[tex]\frac{-x+2}{2}[/tex]

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@Eigenvalue,

If you answer all the chapter test questions, then I get zero practice. Do you agree?

If you answer all the chapter test questions, then no one else here will make an effort to do some work for themselves. Do you agree?

[nycmath]

Question 6

List the intercepts

[tex]x^{2 }[/tex] + y = 9

Let x = 0

[tex]0^{2 }[/tex] + y = 9

0 + y = 9

y = 9

Answer: (0, 9)

Let y = 0

[tex]x^{2 }[/tex] + (0) = 9

[tex]x^{2 }[/tex] = 9

Taking the square on both sides we get x = - and x = 3.

Answer: (-3, 0) and (3, 0)

Test for symmetry

Let y = - y

[tex]x^{2 }[/tex] - y = 9

Let x = - x

[tex](-x)^{2 }[/tex] + y = 9

[tex]x^{2 }[/tex] + y = 9

Let x = - x and y = - y

[tex](-x)^{2 }[/tex] - y = 0

[tex]x^{2 }[/tex] - 9 = 0

The graph of the equation is symmetric about the y-axis.

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Question 7

Use the point-slope formula.

y - [tex]y_{1 }[/tex] = m(x - [tex]x_{1 }[/tex])

y - (-4) = -2(x - 3)

y + 4 = -2x + 6

y = -2x + 6 - 4

y = -2x + 2

The equation has been written in the form y = mx + b, the slope-intercept for.

See attachment for graph.

The graph posted represents the equation y = -2x + 2.

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Question 9...Part 1

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Question 9...Part 2

This is the graph of [tex]x^{2 }[/tex] + [tex]y^{2 }[/tex] + 4x - 2y - 4 = 0.

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Question 10...Part A

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Question 10...Part B

This completes the Chapter 1 Test.