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Sagot :
Given:
The given function is:
[tex]f(x)=-(x-2)^2+9[/tex]
To find:
The transformations, intercepts and the vertex.
Solution:
The vertex form of a parabola is:
[tex]y=a(x-h)^2+k[/tex] ...(i)
Where, a is a constant and (h,k) is vertex.
If a<0, then the graph of parent quadratic function [tex]y=x^2[/tex] reflect across the x-axis.
If h<0, then the graph of parent function shifts h units left and if h>0, then the graph of parent function shifts h units right.
If k<0, then the graph of parent function shifts k units down and if k>0, then the graph of parent function shifts k units up.
We have,
[tex]f(x)=-(x-2)^2+9[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]a=-1,h=2,k=9[/tex]
So, the graph of the parent function reflected across the x-axis, and shifts 2 units right and 9 units up.
Putting x=0 in (ii), we get
[tex]f(0)=-(0-2)^2+9[/tex]
[tex]f(0)=-4+9[/tex]
[tex]f(0)=5[/tex]
The y-intercept is 5.
Putting f(x)=0 in (ii), we get
[tex]0=-(x-2)^2+9[/tex]
[tex](x-2)^2=9[/tex]
Taking square root on both sides, we get
[tex](x-2)=\pm \sqrt{9}[/tex]
[tex]x=\pm 3+2[/tex]
[tex]x=3+2\text{ and }x=-3+2[/tex]
[tex]x=5\text{ and }x=-1[/tex]
Therefore, the x-intercepts are -1 and 5.
The values of h and k are 2 and 9 respectively and (h,k) is the vertex of the parabola.
Therefore, the vertex of the parabola is (2,9).
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