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Chapter 7: Problem 40

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y+4)^{2}=12(x+2)$$

Short Answer

Expert verified
Vertex is (-2, -4), focus is (-2, -1), directrix is \(y = -7\). When graphed, the parabola opens to the right.

Step by step solution

01

Identify the vertex

The vertex of the parabola is given by \((H, K)\) where H and K are the coefficients of x and y in the equation. Thus, from the equation \((y+4)^2 = 12(x+2)\), the vertex is \((-2, -4)\)
02

Find the value of p

The value of p can be determined from the coefficient on the \(x\) term in the equation. Equating the equation to the general form, we can find \(p\). Our given equation is \((y+4)^2 = 12(x+2)\), comparing it with the general equation we get \(4p = 12\), so \(p = 3\)
03

Find the focus

The focus is found at the point \((H, K+p)\). Substituting the known values from the previous steps, the focus comes out to be \((-2, -4+3) = (-2, -1)\)
04

Identify the directrix

The equation of the directrix is given by \(y = K - p\). By substituting the known values in these expressions, the directrix comes out to be \(y = -4 - 3 = -7\)
05

Graph the parabola

On the coordinate plane, plot the vertex point (-2, -4), the focus point (-2, -1) and the directrix line (\(y = -7\)). The parabola will be a curved line that is mirror-symmetrical with the vertex at the peak and the focus inside it. The parabola will be opening to the right side because the x term is isolated

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