The vertex form of the equation of a parabola can be found by?

Question

A. solving for the slope.
B. taking the log.
C. estimating the solutions.
D. dividing by the denominator.
E. completing the square.

Answer 1

Definitely E = completing the square.

Then you can get it in vertex form:
y = a(x-h)² + k

Normally you are given a quadratic equation in the form:
y = ax² + bx + c

So you need to massage it into the vertex form by completing the square. Here's an example:
y = 2x² - 4x + 3

To put this in vertex form you first divide by the coefficent in front of the x² term:
y/2 = x² - 2x + 1½

Now just look at x² - 2x. To complete the square, take the coefficient on the x term (-2), divide it in half (-1) and square it (1). So add this and subtract it from the right side:
y/2 = x² - 2x + 1 - 1 + 1½

Group the first 3 items and see you have a perfect square:
y/2 = (x² - 2x + 1) - 1 + 1½
y/2 = (x - 1)² - 1 + 1½

Simplify the numbers on the right:
y/2 = (x-1)² + ½

Now multiply the 2 back through:
y = 2(x-1)² + 1

There's your equation in vertex form, and it was done by *completing the square* (answer E).

Answer 2

Vertex (- a million, 6): h = - a million ok = 6 x = 0 y = 4 y = a(x - h)² + ok 4 = a[0 - (- a million)]² + 6 4 = a(a million)² + 6 4 = a + 6 a = 4 - 6 a = - 2 Equation: y = - 2(x + a million)² + 6 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯  

Answer 3

E would be the best answer because when you complete the square you get it into the vertex form y=a(x-h)^2 + k.

Answer 4

the answer is e. by completing the square.

ex. y=4ax^2, x=4ay^2
that is a general eq of parabola
when its vertex is at (0,0)

ex
(y-y1)=4a (x-x1)^2

where y1 and y2 is the vertex

u should complete the square in order to have a standard eq or general equation.

hope this helps