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iVinArrow [24]
3 years ago
5

I need the answers to this please and thank you:)

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
6 0

                                                      Q # 1    

Explanation

Given the parabola

 f\left(x\right)=\left(x-3\right)^2-1

Openness

  • It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}

\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c

y=x^2-6x+8

a=1,\:b=-6,\:c=8

x_v=-\frac{\left(-6\right)}{2\cdot \:1}

x_v=3

Finding y_v

y_v=3^2-6\cdot \:3+8

y_v=-1

So vertex is:

\left(3,\:-1\right)

Horizontal Translation

y=\left(x-3\right)^2 moves the graph RIGHT 3 units.

Vertical Translation

 f\left(x\right)=\left(x-3\right)^2-1 moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As a = 1, so it does not affect the stretchiness or compression.

                                       Q # 2  

Explanation:

f\left(x\right)=-\left(x+1\right)^2-2

Openness

  • It OPENS DOWN, as 'a=-1' is negative.

Vertex

\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c

y=-x^2-2x-3

a=-1,\:b=-2,\:c=-3

x_v=-\frac{\left(-2\right)}{2\left(-1\right)}

x_v=-1

\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-2

So vertex is:

\left(-1,\:-2\right)

Horizontal Translation

y=\left(x+1\right)^2 moves the graph LEFT 1 unit.

Vertical Translation

f\left(x\right)=\left(x+1\right)^2-2   moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As a = -1 < 0, so it is either stretched or compressed.

                                          Q # 3  

Explanation:

f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6

It OPENS UP, as 'a=1/3' is positive.

Vertex

\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c

y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}

a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}

x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}

x_v=4            

Finding y_v

y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}

y_v=6            

So vertex is:

\left(4,\:6\right)

Horizontal Translation

f\left(x\right)=\left(x-4\right)^2 moves the graph RIGHT 4 units.

Vertical Translation

f\left(x\right)=}\left(x-4\right)^2+6   moves the graph UP 6 unit.

Stretch or Compress Vertically

As a=\frac{1}{3}, so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

                                             Q # 4

Explanation:

Given the function

 f\left(x\right)=-\left(x+3\right)^2

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form y=a\left(x-m\right)\left(x-n\right)

is the average of the zeros x_v=\frac{m+n}{2}

y=-\left(x+3\right)^2

a=-1,\:m=-3,\:n=-3

x_v=\frac{m+n}{2}

x_v=\frac{\left(-3\right)+\left(-3\right)}{2}

x_v=-3

Finding y_v

y_v=-\left(-3+3\right)^2

y_v=0

So vertex is:

\left(-3,\:0\right)

Horizontal Translation

y=\left(x+3\right)^2 moves the graph LEFT 3 units.

Vertical Translation

y=\left(x+3\right)^2 does not move the graph vertically.

Stretch or Compress Vertically

As a=-1, so it the graph is either vertically stretched or compressed.

                                             Q # 5  

Explanation:

f\left(x\right)=\left(x+5\right)^2-3

Openness

  • It OPENS UP, as 'a=1' is positive.

Vertex

\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c

y=x^2+10x+22

a=1,\:b=10,\:c=22

x_v=-\frac{10}{2\cdot \:1}

x_v=-5

Finding y_v

y_v=\left(-5\right)^2+10\left(-5\right)+22

So vertex is:

\left(-5,\:-3\right)

Horizontal Translation

f\left(x\right)=\left(x+5\right)^2 moves the graph LEFT 5 units.

Vertical Translation

f\left(x\right)=\left(x+5\right)^2-3   moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As a = 1, so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

                                 

                                        Q # 6

THE DETAILS OF COMPLETE SOLUTION OF QUESTION 6 IS ATTACHED IN THE DIAGRAM AS THE 5000 CHARACTERS WERE ALREADY FILLED. SO, I solved via the attached figure.

SO, PLEASE CHECK THE LAST FIGURE TO FIND THE COMPLETE SOLUTION OF THE Q#6.

       

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