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Sagot :
Given:
Consider the given function is
[tex]y=|8x-3|-3[/tex]
To find:
The vertex , axis of symmetry, and transformations of the parent function?
Solution:
We have,
[tex]y=|8x-3|-3[/tex]
[tex]y=\left|8\left(x-\dfrac{3}{8}\right)\right|-3[/tex]
[tex]y=8\left|x-\dfrac{3}{8}\right|-3[/tex] ...(i)
It is an absolute function.
The vertex form of an absolute function is
[tex]y=a|x-h|+k[/tex] ...(ii)
where, a is a constant, (h,k) is vertex and x=h is axis of symmetry.
From (i) and (ii), we get
[tex]a=8,h=\dfrac{3}{8},k=-3[/tex]
So,
[tex]\text{Vertex}:(h,k)=\left(\dfrac{3}{8},-3\right)[/tex]
[tex]\text{Axis of symmetry}:x=\dfrac{3}{8}[/tex]
Parent function of an absolute function is
[tex]y=|x|[/tex]
Since, a=8 therefore, parent function vertically stretched by factor 8.
[tex]h=\dfrac{3}{8}>0[/tex], so the function shifts [tex]\dfrac{3}{8}[/tex] unit right.
k=-3<0, so the function shifts 3 units down.
Therefore, the vertex is [tex]\left(\dfrac{3}{8},-3\right)[/tex] and Axis of symmetry is [tex]x=\dfrac{3}{8}[/tex]. The parent function vertically stretched by factor 8, shifts [tex]\dfrac{3}{8}[/tex] unit right and 3 units down.
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