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Hello, I am currently very stuck with this problem and I am unsure as to how I would solve it.

Hello, I am currently very stuck with this problem and I am unsure as to how I would-example-1
User Zbinsd
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1 Answer

4 votes

We have the equation


20y=x^2-10-15

Let's complete the square, to do it let's add and subtract 25 on the right side


\begin{gathered} 20y=x^2-10-15+25-25 \\ \\ 20y=(x-5)^2-15-25_{} \\ \\ 20y=(x-5)^2-40 \\ \\ \end{gathered}

Now we can have y in function of x


\begin{gathered} y=(1)/(20)(x-5)^2-2 \\ \\ \end{gathered}

Now we can already identify the vertex because it's in the vertex form:


y=a(x-h)+k

Where the vertex is


(h,k)

As we can see, h = 5 and k = -2, then the vertex is


(5,-2)

Now we can continue and find the focus, the focus is


\mleft(h,k+(1)/(4a)\mright)

We have a = 1/20, therefore


\begin{gathered} \mleft(5,-2+5\mright) \\ \\ (5,3) \end{gathered}

The focus is


(5,3)

And the last one, the directrix, it's


y=k-(1)/(4a)

Then


\begin{gathered} y=-2-5 \\ \\ y=-7 \end{gathered}

Hence the correct answer is: vertex (5, -2); focus (5, 3); directrix y = -7

User Todd Sjolander
by
9.1k points

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