What is the focus of the parabola y = -8x^2 - 5? Simplify any fractions. (______,______)

Identify vertex form: Identify the vertex form of the parabola. The given equation y = -8x^2 - 5 is similar to y = a(x - h)^2 + k, where a = -8, h = 0, and k = -5.Calculate p value: Calculate the value of p using the formula p = 1/(4a). p = 1/(4*(-8)) p = -1/32Find focus: Find the focus of the parabola using the vertex (h, k) and the value of p. The focus is at (h, k + p). Focus = (0, -5 - 1/32) Focus = (0, -160/32 - 1/32) Focus = (0, -161/32)

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What is the focus of the parabola $y = -8x^2 - 5$ ? $\newline$ Simplify any fractions. $\newline$ (,)

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Algebra 2

Find the focus or directrix of a parabola

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Q. What is the focus of the parabola

y = -8x^2 - 5

\newline

Simplify any fractions.

\newline

(______,______)

Identify vertex form: Identify the vertex form of the parabola. $\newline$ The given equation $y = -8x^2 - 5$ is similar to $y = a(x - h)^2 + k$ , where $a = -8$ , $h = 0$ , and $k = -5$ .
Calculate p value: Calculate the value of $p$ using the formula $p = \frac{1}{4a}$ . $p = \frac{1}{4(-8)}$ $p = -\frac{1}{32}$
Find focus: Find the focus of the parabola using the vertex $(h, k)$ and the value of $p$ . The focus is at $(h, k + p)$ . Focus = $(0, -5 - \frac{1}{32})$ Focus = $(0, -\frac{160}{32} - \frac{1}{32})$ Focus = $(0, -\frac{161}{32})$