name the coordinates of the vertex of the graph of y=2x^2+4x-1...?

Answer 1

note
ax^2 + bx + c is the generic equation of a quadratic

The x value of the vertex is found at

-b/(2a) = -4/(2*2) = -1

to find the y value plug it into your equation, so

y = 2(-1)^2 + 4(-1) - 1
y = 2(1) - 4 - 1
y = -3

Vertex (-1, -3)

Answer 2

a = 1, b = 4, c = 1. The axis of symmetry exists at x = -b/2a which is -4/2 = -2. This is the x coord of the vertex. Find f(-2) to find the y-value of the vertex. f(-2) = 4 - 8 + 1 = -3. The vertex is at (-2, -3). The y-intercept is at (0, 1). You can find a third point by reflecting (0, 1) across x = -2. This gives you (-4, 1). From the vertex go over +/-1, up 1 to get (-1, -2) and (-3, -2). Connect these 5 points to get a beautiful, perfect parabola!

Answer 3

this is a parabola opening upward.

convert it to std. eq.

(x -h)^2 = 4a(y -k), bec the axis parallel to the y-axis

where (h, k) is the vertex

y = 2x^2 + 4x - 1, subtract both sides by 1

y +1 = 2x^2 + 4x, divide both sides by 2

(y + 1)/2 = x^2 + 2x, then add 1 to both sides

1/2 (y + 1) + 1 = x^2 + 2x + 1, simplify

1/2 (y + 3) = (x + 1)^2, then we get

(x + 1)^2 = 1/2 (y+3), from here we can get h = -1, k = -3

therefore our vertex is @ (-1, -3)

Answer 4

Use completing the square to get it in the proper form:

y + 1 = 2x^2 + 4x
y + 1 = 2(x^2 + 2x)
y + 1 = 2(x^2 + 2x + 1) - 2
y + 3 = 2(x^2 + 2x + 1)
(y + 3) = 2(x + 1)^2

The vertex is at (-1, -3)

Answer 5

To find the x coordinate of the vertex of a parabola, use
-b/2a. To find the y coordinate of the vertex of a parabola, use (-b^2/4a)+c.
So the x coordinate is -1
The y coordinate is -3
(-1,-3)

Answer 6

f `(x) = 4x + 4 = 0 for vertex
4x = - 4
x = - 1
f (-1) = 2 - 4 - 1 = - 3
Vertex is V(-1,-3)