What is the focus of the parabola y = 6x^2 - 3? Simplify any fractions. (______,______)

Identify Parabola Form: y = 6x^2 - 3 is a vertical parabola cuz it's in the form y = ax^2 + bx + c.Find Vertex Form: The vertex form of a parabola is y = a(x - h)^2 + k, so we need to find a, h, and k.Determine Parameters: Comparing y = 6x^2 - 3 to y = a(x - h)^2 + k, we get a = 6, h = 0, and k = -3.Calculate p Value: Now we find p using the formula p = 1/(4a). So p = 1/(4*6) = 1/24.Locate Focus: The focus of a vertical parabola is (h, k + p). So we plug in h = 0, k = -3, and p = 1/24.Simplify Coordinates: The focus is (0, -3 + 1/24). We simplify -3 + 1/24 to get -72/24 + 1/24.After simplifying, we get -71/24. So the focus is (0, -71/24).

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What is the focus of the parabola $y = 6x^2 - 3$ ? $\newline$ 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 = 6x^2 - 3

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Simplify any fractions.

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(______,______)

Identify Parabola Form: $y = 6x^2 - 3$ is a vertical parabola cuz it's in the form $y = ax^2 + bx + c$ .
Find Vertex Form: The vertex form of a parabola is $y = a(x - h)^2 + k$ , so we need to find $a$ , $h$ , and $k$ .
Determine Parameters: Comparing $y = 6x^2 - 3$ to $y = a(x - h)^2 + k$ , we get $a = 6$ , $h = 0$ , and $k = -3$ .
Calculate p Value: Now we find $p$ using the formula $p = \frac{1}{4a}$ . So $p = \frac{1}{4\times 6} = \frac{1}{24}$ .
Locate Focus: The focus of a vertical parabola is $(h, k + p)$ . So we plug in $h = 0$ , $k = -3$ , and $p = \frac{1}{24}$ .
Simplify Coordinates: The focus is $(0, -3 + 1/24)$ . We simplify $-3 + 1/24$ to get $-72/24 + 1/24$ .
Simplify Coordinates: The focus is (0, -3 + rac{1}{24}). We simplify -3 + rac{1}{24} to get -rac{72}{24} + rac{1}{24}.After simplifying, we get -rac{71}{24}. So the focus is (0, -rac{71}{24}).