Question

Match each quadratic function with the attributes of its parabolic graph.

Answer 1

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:

$f(x)=-0.3(x-2)^2+5$

The standard equation of an up-facing parabola with a vertex at (h,k) and a focal length |p| is given as:

$(x-h)^2=4p(y-k)$

We rewrite the given equation in the form above:

$\begin{gathered} f(x)=-0.3(x-2)^2+5 \\ f(x)-5=-\frac{3}{10}(x-2)^2 \\ -\frac{10}{3}[f(x)-5]=(x-2)^2 \\ \left(x-2\right)^2=4(-\frac{5}{6})[f(x)-5] \end{gathered}$

From the form above:

• Option A: ,The vertex, (h,k) = (2, 5)

,

• Option E: ,Axis of Symmetry, x=2

$Focus,(h,k+p)=(2,5-\frac{5}{6})=(2,4\frac{1}{6})$

• Option I: ,The focus is (2, 4 1/6)

Part B

Given the equation:

$f(x)=0.2(x+2)^2-5$

Rewrite the equation in the standard form given earlier:

$\begin{gathered} 0.2(x+2)^2=f(x)+5 \\ (x+2)^2=5[f(x)+5] \\ (x+2)^2=4(\frac{5}{4})[f(x)+5] \end{gathered}$

From the form above:

• Option C: ,The vertex, (h,k) = (-2, -5)

,

• Option F: ,Axis of Symmetry, x= -2

$Focus,(h,k+p)=(-2,-5+\frac{5}{4})=(-2,-3\frac{3}{4})$

• Option G: ,The focus is (-2, 3 3/4)