Przesuwając wykres funkcji [tex]f(x)[/tex] o wektor [tex]\overrightarrow{u}=[p,q][/tex], otrzymujemy wykres funkcji
[tex]g(x)=f(x-p)+q[/tex]
a)
[tex]f(x)=3x^2\qquad\qquad \overrightarrow{u}=[2,3]\\\\g(x)=f(x-2)+3=3(x-2)^2+3=3(x^2-4x+4)+3=3x^2-12x+12+3=\\\\=3x^2-12x+15[/tex]
b)
[tex]f(x)=\frac{1}{2}x^2\qquad\qquad \overrightarrow{u}=[-1,0]\\\\g(x)=f(x+1)+0=f(x+1)=\frac{1}{2}(x+1)^2=\frac{1}{2}(x^2+2x+1)=\frac{1}{2}x^2+x+\frac{1}{2}[/tex]
c)
[tex]f(x)=-0,1x^2\qquad\qquad \overrightarrow{u}=[0,-4]\\\\g(x)=f(x-0)-4=f(x)-4=-0,1x^2-4[/tex]
d)
[tex]f(x)=\frac{3}{4}x^2\qquad\qquad \overrightarrow{u}=[-3,-7]\\\\g(x)=f(x+3)-7=\frac{3}{4}(x+3)^2-7=\frac{3}{4}(x^2+6x+9)-7=\frac{3}{4}x^2+\frac{18}{4}x+\frac{27}{4}-7=\\\\=\frac{3}{4}x^2+\frac{9}{2}x+6\frac{3}{4}-7=\frac{3}{4}x^2+4\frac{1}{2}x-\frac{1}{4}[/tex]
e)
[tex]f(x)=-x^2\qquad\qquad \overrightarrow{u}=[2,-5]\\\\g(x)=f(x-2)-5=-(x-2)^2-5=-(x^2-4x+4)-5=-x^2+4x-4-5=\\\\=-x^2+4x-9[/tex]
f)
[tex]f(x)=\frac{x^2}{4}\qquad\qquad \overrightarrow{u}=[-6,-10]\\\\g(x)=f(x+6)-10=\frac{(x+6)^2}{4}-10=\frac{x^2+12x+36}{4}-10=\frac{1}{4}x^2+3x+9-10=\\\\=\frac{1}{4}x^2+3x-1[/tex]
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Przesuwając wykres funkcji [tex]f(x)[/tex] o wektor [tex]\overrightarrow{u}=[p,q][/tex], otrzymujemy wykres funkcji
[tex]g(x)=f(x-p)+q[/tex]
a)
[tex]f(x)=3x^2\qquad\qquad \overrightarrow{u}=[2,3]\\\\g(x)=f(x-2)+3=3(x-2)^2+3=3(x^2-4x+4)+3=3x^2-12x+12+3=\\\\=3x^2-12x+15[/tex]
b)
[tex]f(x)=\frac{1}{2}x^2\qquad\qquad \overrightarrow{u}=[-1,0]\\\\g(x)=f(x+1)+0=f(x+1)=\frac{1}{2}(x+1)^2=\frac{1}{2}(x^2+2x+1)=\frac{1}{2}x^2+x+\frac{1}{2}[/tex]
c)
[tex]f(x)=-0,1x^2\qquad\qquad \overrightarrow{u}=[0,-4]\\\\g(x)=f(x-0)-4=f(x)-4=-0,1x^2-4[/tex]
d)
[tex]f(x)=\frac{3}{4}x^2\qquad\qquad \overrightarrow{u}=[-3,-7]\\\\g(x)=f(x+3)-7=\frac{3}{4}(x+3)^2-7=\frac{3}{4}(x^2+6x+9)-7=\frac{3}{4}x^2+\frac{18}{4}x+\frac{27}{4}-7=\\\\=\frac{3}{4}x^2+\frac{9}{2}x+6\frac{3}{4}-7=\frac{3}{4}x^2+4\frac{1}{2}x-\frac{1}{4}[/tex]
e)
[tex]f(x)=-x^2\qquad\qquad \overrightarrow{u}=[2,-5]\\\\g(x)=f(x-2)-5=-(x-2)^2-5=-(x^2-4x+4)-5=-x^2+4x-4-5=\\\\=-x^2+4x-9[/tex]
f)
[tex]f(x)=\frac{x^2}{4}\qquad\qquad \overrightarrow{u}=[-6,-10]\\\\g(x)=f(x+6)-10=\frac{(x+6)^2}{4}-10=\frac{x^2+12x+36}{4}-10=\frac{1}{4}x^2+3x+9-10=\\\\=\frac{1}{4}x^2+3x-1[/tex]