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Chapter 7: Problem 36

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(x+2)^{2}=4(y+1)$$

Short Answer

Expert verified
The vertex of the parabola is (-2, -1), the focus is (-2, 0), and the equation of the directrix is \(y = -2\).

Step by step solution

01

Identify a, h and k

From the given equation, we can observe that the parabola is in the form of \( (x - h)^2 = 4a(y - k)\). By comparing the equation \((x+2)^2=4(y+1)\) with the standard form, we get \(h = -2, k = -1\) and \( a = 1\).
02

Compute the vertex

The vertex of the parabola is given as (h, k). So, by substituting the values we just found, we obtain the vertex, (-2,-1).
03

Compute the focus

The focus of the parabola is given as (h, k + a). We substitute h = -2, k = -1 and a = 1 into our equation and it gives us our focus, (-2, 0).
04

Compute the directrix

The directrix of the parabola is given by the equation \(y = k - a\). Substituting k = -1 and a = 1 into our equation gives the equation of the directrix, \(y = -2\)
05

Graph the parabola

On the Cartesian plane, plot the vertex at (-2,-1), then it is known that the parabola opens upwards because a > 0. Plot the focus at (-2, 0) and draw the directrix at \(y = -2\). Your parabola should be symmetric with respect to the line x = -2, which is its axis of symmetry.

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