Question

graph the following parabola by moving the vertex and an endpoint of the latus rectum. $(y + 1)^2 = 4x$ provide your answer below:

Explanation:

Step1: Identify the vertex form

The standard - form of a parabola opening to the right or left is \((y - k)^2=4p(x - h)\), where \((h,k)\) is the vertex. For the equation \((y + 1)^2 = 4x\), we have \(h = 0\), \(k=-1\), so the vertex is \((0,-1)\).

Step2: Find the value of \(p\)

Comparing \((y + 1)^2 = 4x\) with \((y - k)^2=4p(x - h)\), we get \(4p = 4\), so \(p = 1\).

Step3: Determine the endpoints of the latus - rectum

The endpoints of the latus - rectum of a parabola \((y - k)^2=4p(x - h)\) are \((h + p,k\pm2p)\). Substituting \(h = 0\), \(k=-1\), and \(p = 1\), we get the endpoints \((1,1)\) and \((1,-3)\).

Step4: Graph the parabola

Plot the vertex \((0,-1)\) and the endpoints of the latus - rectum \((1,1)\) and \((1,-3)\) and then draw a smooth curve passing through these points.

Answer:

The vertex of the parabola \((y + 1)^2 = 4x\) is \((0,-1)\), and the endpoints of the latus - rectum are \((1,1)\) and \((1,-3)\). Graph by plotting these points and drawing a smooth curve.