Answer:
f(x)=-0.3(x-2)²+5
• Option A: ,The vertex, (h,k) = (2, 5)
,• Option E: ,Axis of Symmetry, x=2
• Option I: ,The focus is (2, 4 1/6)
f(x)=0.2(x+2)²-5
• Option C: ,The vertex, (h,k) = (-2, -5)
,• Option F: ,Axis of Symmetry, x= -2
• Option G: ,The focus is (-2, 3 3/4)
Explanation:
Part A
Given the equation:
[tex]f(x)=-0.3(x-2)^2+5[/tex]The standard equation of an up-facing parabola with a vertex at (h,k) and a focal length |p| is given as:
[tex](x-h)^2=4p(y-k)[/tex]We rewrite the given equation in the form above:
[tex]\begin{gathered} f(x)=-0.3(x-2)^2+5 \\ f(x)-5=-\frac{3}{10}(x-2)^2 \\ -\frac{10}{3}[f(x)-5]=(x-2)^2 \\ \left(x-2\right)^2=4(-\frac{5}{6})[f(x)-5] \end{gathered}[/tex]From the form above:
• Option A: ,The vertex, (h,k) = (2, 5)
,• Option E: ,Axis of Symmetry, x=2
[tex]Focus,(h,k+p)=(2,5-\frac{5}{6})=(2,4\frac{1}{6})[/tex]• Option I: ,The focus is (2, 4 1/6)
Part B
Given the equation:
[tex]f(x)=0.2(x+2)^2-5[/tex]Rewrite the equation in the standard form given earlier:
[tex]\begin{gathered} 0.2(x+2)^2=f(x)+5 \\ (x+2)^2=5[f(x)+5] \\ (x+2)^2=4(\frac{5}{4})[f(x)+5] \end{gathered}[/tex]From the form above:
• Option C: ,The vertex, (h,k) = (-2, -5)
,• Option F: ,Axis of Symmetry, x= -2
[tex]Focus,(h,k+p)=(-2,-5+\frac{5}{4})=(-2,-3\frac{3}{4})[/tex]• Option G: ,The focus is (-2, 3 3/4)
