Question

Get the parabola in the appropriate form so you can identify the following:-8y=x^2+8x-8Opens:P:Vertex: Focus:Directrix:Axis of Symmetry:

Answer 1

Given

`-8y=x^2+8x-8`

Solving for y,

`\Rightarrow y=-(1)/(8)x^2-x+1`

Therefore, since the coefficient that accompanies the x^2 is negative, the parabola opens downwards.

In general, the vertex and general form of a parabola are

`\begin{gathered} y=a(x-h)^2+k\to\text{vertex form} \\ y=ax^2+bx+c \end{gathered}`

Where (h,k) is the vertex of the parabola

Then,

`\begin{gathered} \Rightarrow ax^2+bx+c=a(x-h)^2+k=ax^2-2ahx+ah^2+k \\ \Rightarrow\begin{cases}a=a \\ b=-2ah \\ c=ah^2+k\end{cases} \end{gathered}`

Finding h and k in our case,

`\begin{gathered} \Rightarrow-1=-2(-(1)/(8))h\Rightarrow h=-4 \\ \Rightarrow1=-(1)/(8)(-4)^2+k \\ \Rightarrow k=1+2 \\ \Rightarrow k=3 \\ \Rightarrow(h,k)=(-4,3) \end{gathered}`

The vertex of the parabola is (-4,3)

Notice that the parabola opens downwards and it is parallel to the y-axis; therefore,

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

Where (h,k) is the vertex. In our case, from the vertex form of the quadratic equation,

`\begin{gathered} y=-(1)/(8)(x+4)^2+3 \\ \Rightarrow(x+4)^2=-8(y-3) \end{gathered}`

Thus,

`\begin{gathered} \Rightarrow4p=-8 \\ \Rightarrow p=-2 \end{gathered}`

The distance from the focus to the vertex is 2.

Because the graph of the function opens downwards, we can find its focus as shown below,

`\begin{gathered} (h,k)=(-4,3) \\ \Rightarrow focus\colon(-4,3-2)=(-4,1)_{} \\ \Rightarrow\text{focus(-4,1)} \end{gathered}`

The focus is (-4,1)

Because the graph opens downward and p=-2, the equation of the directrix is

`\begin{gathered} y=k+|p|=3+2 \\ \Rightarrow y=5 \end{gathered}`

The directrix is y=5

Finally, the axis of symmetry is a line perpendicular to the directrix and crosses the vertex; therefore,

`\begin{gathered} y=5\to\text{ slope=0} \\ (h,k)=(-4,3) \\ \Rightarrow x=-4\to\text{ axis of sy}mmetry \end{gathered}`

The axis of symmetry is x=-4