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I just need help with question one, but if you want to you can answer question 2 as well. I’ll give 100 points!

I Just Need Help With Question One But If You Want To You Can Answer Question 2 As Well Ill Give 100 Points class=

Sagot :

fieryanswererft

Explanation:

Given f(x) : (2, -3)

Translation's:

f(x) + 2 then graph translates up by 2 units up =  [tex]\boxed{\sf (2, -1)}[/tex]

f(x) - 3 then graph translates down 3 units down = [tex]\boxed{\sf (2, -6)}[/tex]

f(x + 5) then graph translates left 5 units = [tex]\boxed{\sf (-3, -3)}[/tex]

-f(x) then graph reflects over x axis = [tex]\sf \boxed{\sf (2, 3)}[/tex]

f(-x)  then graph reflects over y axis = [tex]\sf \boxed{\sf (-2,-3)}[/tex]

f(2x) then graph has horizontal compression = (2/2, -3) = [tex]\boxed{\sf (1, -3)}[/tex]

2f(x) then graph has vertical compression = (2, (-3)2) = [tex]\boxed{\sf (2, -6)}[/tex]

-f(x - 4) then graph reflects over x axis, moves 4 units to right = [tex]\sf \boxed{\sf (6, 3)}[/tex]

Solution 2

Parent function: y = x²

Graph function: f(x) = (x + 8)² - 4

After Identification:

D. The graph has a translation of 8 units left and 4 units down.

semsee45

Answer:

Translations

For [tex]a > 0[/tex]

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}[/tex]

[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]

[tex]y=f(ax) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{a}[/tex]

[tex]y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}[/tex]

[tex]y=f(-x) \implies f(x) \: \textsf{reflected in the} \: y \textsf{-axis}[/tex]

Question 1

Given:  [tex]f(x)=(2,-3)[/tex]

[tex]f(x)+2 \implies (x, y+2)= (2,-3+2)=(2,-1)[/tex]

[tex]f(x)-3 \implies (x,y-3)=(2,-3-3)=(2,-6)[/tex]

[tex]f(x+5)\implies (x-5,y)=(2-5,-3)=(-3,-3)[/tex]

[tex]-f(x) \implies (x,-y)=(2,-(-3))=(2,3)[/tex]

[tex]f(-x) \implies (-x,y)=(-(2),-3)=(-2,-3)[/tex]

[tex]f(2x) \implies \left(\dfrac{x}{2},y\right)=\left(\dfrac{2}{2},-3\right)=(1,-3)[/tex]

[tex]2f(x) \implies (x,2y)=(2,2 \cdot -3)=(2,-6)[/tex]

[tex]-f(x-4) \implies (x+4,-y)=(2+4,-(-3))=(6,3)[/tex]

[tex]\begin{array}{| c | c | c | c | c | c | c | c |}\cline{1-8} & & & & & & &\\f(x)+2 & f(x)-3 & f(x+5) & -f(x) & f(-x) & f(2x) & 2f(x) & -f(x-4)\\& & & & & & &\\\cline{1-8} & & & & & & &\\(2,-1) & (2,-6) & (-3,-3) & (2,3) & (-2,-3) & (1,-3) & (2,-6) & (6,3)\\& & & & & & &\\\cline{1-8} \end{array}[/tex]

Question 2

Parent function:  [tex]y=x^2[/tex]

Given function:  [tex]f(x)=(x+8)^2-4[/tex]

[tex]f(x+8) \implies f(x) \: \textsf{translated}\:8\:\textsf{units left}[/tex]

[tex]f(x)-4 \implies f(x) \: \textsf{translated}\:4\:\textsf{units down}[/tex]

Therefore, a translation 8 units to the left and 4 units down.

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