Q # 1
Explanation
Given the parabola
[tex]f\left(x\right)=\left(x-3\right)^2-1[/tex]
Openness
- It OPENS UP, as 'a=1' is positive.
Finding Vertex
The vertex of an up-down facing parabola of the form
[tex]y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}[/tex]
[tex]\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c[/tex]
[tex]y=x^2-6x+8[/tex]
[tex]a=1,\:b=-6,\:c=8[/tex]
[tex]x_v=-\frac{\left(-6\right)}{2\cdot \:1}[/tex]
[tex]x_v=3[/tex]
Finding [tex]y_v[/tex]
[tex]y_v=3^2-6\cdot \:3+8[/tex]
[tex]y_v=-1[/tex]
So vertex is:
[tex]\left(3,\:-1\right)[/tex]
Horizontal Translation
[tex]y=\left(x-3\right)^2[/tex] moves the graph RIGHT 3 units.
Vertical Translation
[tex]f\left(x\right)=\left(x-3\right)^2-1[/tex] moves the graph DOWN 1 unit.
Stretch or Compress Vertically
As [tex]a = 1[/tex], so it does not affect the stretchiness or compression.
Q # 2
Explanation:
[tex]f\left(x\right)=-\left(x+1\right)^2-2[/tex]
Openness
- It OPENS DOWN, as 'a=-1' is negative.
Vertex
[tex]\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c[/tex]
[tex]y=-x^2-2x-3[/tex]
[tex]a=-1,\:b=-2,\:c=-3[/tex]
[tex]x_v=-\frac{\left(-2\right)}{2\left(-1\right)}[/tex]
[tex]x_v=-1[/tex]
[tex]\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}[/tex]
[tex]y_v=-2[/tex]
So vertex is:
[tex]\left(-1,\:-2\right)[/tex]
Horizontal Translation
[tex]y=\left(x+1\right)^2[/tex] moves the graph LEFT 1 unit.
Vertical Translation
[tex]f\left(x\right)=\left(x+1\right)^2-2[/tex] moves the graph DOWN 2 unit.
Stretch or Compress Vertically
As [tex]a = -1[/tex] < 0, so it is either stretched or compressed.
Q # 3
Explanation:
[tex]f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6[/tex]
It OPENS UP, as 'a=1/3' is positive.
Vertex
[tex]\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c[/tex]
[tex]y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}[/tex]
[tex]a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}[/tex]
[tex]x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}[/tex]
[tex]x_v=4[/tex]
Finding [tex]y_v[/tex]
[tex]y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}[/tex]
[tex]y_v=6[/tex]
So vertex is:
[tex]\left(4,\:6\right)[/tex]
Horizontal Translation
[tex]f\left(x\right)=\left(x-4\right)^2[/tex] moves the graph RIGHT 4 units.
Vertical Translation
[tex]f\left(x\right)=}\left(x-4\right)^2+6[/tex] moves the graph UP 6 unit.
Stretch or Compress Vertically
As [tex]a=\frac{1}{3}<1[/tex], so it the graph is vertically compressed by a factor of 1/3.
Check the attached comparison graphs.
Q # 4
Explanation:
Given the function
[tex]f\left(x\right)=-\left(x+3\right)^2[/tex]
It OPENS DOWN, as 'a=-1' is negative.
Vertex
The vertex of an up-down facing parabola of the form [tex]y=a\left(x-m\right)\left(x-n\right)[/tex]
is the average of the zeros [tex]x_v=\frac{m+n}{2}[/tex]
[tex]y=-\left(x+3\right)^2[/tex]
[tex]a=-1,\:m=-3,\:n=-3[/tex]
[tex]x_v=\frac{m+n}{2}[/tex]
[tex]x_v=\frac{\left(-3\right)+\left(-3\right)}{2}[/tex]
[tex]x_v=-3[/tex]
Finding [tex]y_v[/tex]
[tex]y_v=-\left(-3+3\right)^2[/tex]
[tex]y_v=0[/tex]
So vertex is:
[tex]\left(-3,\:0\right)[/tex]
Horizontal Translation
[tex]y=\left(x+3\right)^2[/tex] moves the graph LEFT 3 units.
Vertical Translation
[tex]y=\left(x+3\right)^2[/tex] does not move the graph vertically.
Stretch or Compress Vertically
As [tex]a=-1<1[/tex], so it the graph is either vertically stretched or compressed.
Q # 5
Explanation:
[tex]f\left(x\right)=\left(x+5\right)^2-3[/tex]
Openness
- It OPENS UP, as 'a=1' is positive.
Vertex
[tex]\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c[/tex]
[tex]y=x^2+10x+22[/tex]
[tex]a=1,\:b=10,\:c=22[/tex]
[tex]x_v=-\frac{10}{2\cdot \:1}[/tex]
[tex]x_v=-5[/tex]
Finding [tex]y_v[/tex]
[tex]y_v=\left(-5\right)^2+10\left(-5\right)+22[/tex]
So vertex is:
[tex]\left(-5,\:-3\right)[/tex]
Horizontal Translation
[tex]f\left(x\right)=\left(x+5\right)^2[/tex] moves the graph LEFT 5 units.
Vertical Translation
[tex]f\left(x\right)=\left(x+5\right)^2-3[/tex] moves the graph DOWN 3 unit.
Stretch or Compress Vertically
As [tex]a = 1[/tex], 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.
