Question

question
graph the parabola given by the equation (x + 1)^2 = -16(y - 4) by dragging the vertex and the additional point to one end of the latus rectum.
provide your answer below:

Explanation:

Step1: Identify the vertex form of parabola

The general form of a vertical - axis parabola is $(x - h)^2=4p(y - k)$, and the given equation is $(x + 1)^2=-16(y - 4)$. By comparing, we have $h=-1,k = 4$. So the vertex of the parabola is $(-1,4)$.

Step2: Find the value of $p$

Since $4p=-16$, then $p=-4$.

Step3: Find the endpoints of the latus - rectum

The length of the latus - rectum is $|4p| = 16$. For a parabola of the form $(x - h)^2=4p(y - k)$, the endpoints of the latus - rectum have $y=k + p$. Here, $y=4-4 = 0$. Substitute $y = 0$ into the equation $(x + 1)^2=-16(0 - 4)=64$. Then $x+1=\pm8$, so $x=-1\pm8$. The endpoints of the latus - rectum are $(7,0)$ and $(-9,0)$.

To graph: Drag the vertex to the point $(-1,4)$ and an additional point to either $(7,0)$ or $(-9,0)$.

Answer:

Vertex: $(-1,4)$; End - point of latus rectum: $(7,0)$ or $(-9,0)$