What is the focus of the parabola y = -10x^2 - 3? Simplify any fractions. (______,______)

Identify vertex form: Identify the vertex form of the parabola. y = a(x - h)^2 + k y = -10x^2 - 3 can be rewritten as y = -10(x - 0)^2 - 3 So, h = 0 and k = -3.Find value of 'p': Find the value of 'p' using the formula p = 1/(4a). p = 1/(4*(-10)) p = -1/40Determine focus: Determine the focus of the parabola. The focus is (h, k + p). Substitute h = 0, k = -3, and p = -1/40. Focus = (0, -3 - 1/40) Focus = (0, -120/40 - 1/40) Focus = (0, -121/40)

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What is the focus of the parabola $y = -10x^2 - 3$ ? $\newline$ 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 = -10x^2 - 3

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Simplify any fractions.

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(______,______)

Identify vertex form: Identify the vertex form of the parabola. $\newline$ $y = a(x - h)^2 + k$ $\newline$ $y = -10x^2 - 3$ can be rewritten as $y = -10(x - 0)^2 - 3$ $\newline$ So, $h = 0$ and $k = -3$ .
Find value of 'p': Find the value of 'p' using the formula $p = \frac{1}{4a}$ . $p = \frac{1}{4(-10)}$ $p = -\frac{1}{40}$
Determine focus: Determine the focus of the parabola. $\newline$ The focus is $(h, k + p)$ . $\newline$ Substitute $h = 0$ , $k = -3$ , and $p = -\frac{1}{40}$ . $\newline$ Focus = $(0, -3 - \frac{1}{40})$ $\newline$ Focus = $(0, -\frac{120}{40} - \frac{1}{40})$ $\newline$ Focus = $(0, -\frac{121}{40})$