What is the focus of the parabola y = 8x^2 - 4? Simplify any fractions. (______,______)

Identify Parabola Type: We got y = 8x^2 - 4, which is a vertical parabola. Find the vertex form by completing the square if needed. But here, no need to complete the square since x is already squared and there's no x term. So, the vertex form is y = 8(x - 0)^2 - 4, where a = 8, h = 0, and k = -4.Find Vertex Form: Now, let's find the value of p using the formula p = 1/(4a). So, p = 1/(4*8). p = 1/32.Calculate p Value: The focus of a vertical parabola is (h, k + p). We already know h = 0, k = -4, and p = 1/32. So, the focus is (0, -4 + 1/32).Determine Focus Coordinates: Now, let's simplify -4 + 1/32. -4 is the same as -128/32. So, -128/32 + 1/32 = -127/32. The focus is (0, -127/32).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the focus of the parabola $y = 8x^2 - 4$ ? $\newline$ Simplify any fractions. $\newline$ (,)

View step-by-step help

Home

Math Problems

Algebra 2

Find the focus or directrix of a parabola

Full solution

Q. What is the focus of the parabola

y = 8x^2 - 4

\newline

Simplify any fractions.

\newline

(______,______)

Identify Parabola Type: We got $y = 8x^2 - 4$ , which is a vertical parabola. $\newline$ Find the vertex form by completing the square if needed. $\newline$ But here, no need to complete the square since $x$ is already squared and there's no $x$ term. $\newline$ So, the vertex form is $y = 8(x - 0)^2 - 4$ , where $a = 8$ , $h = 0$ , and $k = -4$ .
Find Vertex Form: Now, let's find the value of $p$ using the formula $p = \frac{1}{4a}$ . So, $p = \frac{1}{4\times 8}$ . $p = \frac{1}{32}$ .
Calculate p Value: The focus of a vertical parabola is $(h, k + p)$ . We already know $h = 0$ , $k = -4$ , and $p = \frac{1}{32}$ . So, the focus is $(0, -4 + \frac{1}{32})$ .
Determine Focus Coordinates: Now, let's simplify $-4 + \frac{1}{32}$ . $-4$ is the same as $-\frac{128}{32}$ .So, $-\frac{128}{32} + \frac{1}{32} = -\frac{127}{32}$ .The focus is $(0, -\frac{127}{32})$ .