Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

# How do I find the vertex of this parabola?

How do I find the vertex of this parabola?

y^2 = 16x

Also, any idea on finding the focus of this parabola too?

Relevance

First note that the function is quadratic in y as opposed to x. To get it into the familiar form, just swap x & y. We can do this because this amounts to just rotating the parabola, and it does not affect the geometry of the parabola, itself. Thus, x = (1/16)y^2 is transformed to y = (1/16)x^2. The focus = 1/(4a) = 4, and the (transformed) x-component of the vertex is x = -b/(2a) = 0, and therefore, so does y=0; and the transformed vertex is (0,0). The focus, as a point in the transformed plane, in this case is (0,4). Thus, swapping y and x in order to transform back, we have the vertex (0,0), and focus at (4,0).

• A parabola of the style y-ok = (x-h)^2 or y=(x-h)^2+ok has (h,ok) because of the fact the vertex. In all your issues, the x-coordinate of the vertex is 0 as x^2=(x-0)^2 y=x^2 vertex(0,0) y=x^2+5 vertex (0.5) y=x^2-3 vertex (0,-3) y=x^2-2 vertex (0,-2) to end the sq. x^2+4x, divide the coefficient of x with the aid of 2. (x+2)^2=x^2+4x+4 so, x^2+4x = (x+2)^2-4 (x^2-12x) comparable concept. x^2-12x=(x-6)^2-36

• y² = 16x can be written

x = (1/16)y²

so this is a parabola that opens to the right and has a vertex (0,0).

y = ax² + bx + c ...or... y = a(x-h)² + k

just "swap the x's and y's ...but be careful writing the points as the x values and y values must still be written in their proper order.

good luck

• (y - k)² = 4p(x - h) is the standard form of the equation for a horizontal p'bola with vertex at (h,k) and opening to the right.

y² = 4px is the standard form with vertex at origin . . .

p = distance from vertex to focus

4p = 16, solve for p