Question

HELP!!!

Answer 1

iCan't Even Help You Cause i'm Still In Chapter 4

Answer 2

dang this is my last assignment and then i can take my semester test :( i need this like tonight like before 8:00

Answer 3

woe idk any of that.. good luck tho!

Answer 4

i just dont understand a lot of it because i rushed through it so i can do my semester test tonight.... please help ;/

Answer 5

@Hero can you help me??

Answer 6

Here's a start for understanding quadratic equations in standard form and vertex form.

Standard form: y = ax^2 + bx + c

If "a" is positive, the parabola will open upward and the function will have minimum.
If "a" is negative, the parabola will open downwards and the function will have a maximum.
The function will have 0, 1, or 2 real zeros, values for "x" for which the function will have a "y" value equal to "0". The graph will touch or cross the x-axis at those places (points).

To convert to vertex form:

y = ax^2 + bx + c -> y = a[x^2 + (b/a)x] + c ->
y = a[x^2 + (b/a)x + (b/2a)^2] + c - (b^2)/(4a) ->
y = a[x + (b/2a)]^2 + c - (b^2)/(4a)

So, you have a max or min at (-[b/2a], c - [b^2]/[4a]) That's your vertex.

Answer 7

The axis of symmetry is at x = -(b/2a) and that "x" is the same for your x-coordinate in your vertex. Goes right through the vertex.

Answer 8

The domain will be all values of "x". The range will depend on whether "a" is positive (function has a min, opens upwards, looks like a regular bowl) or negative (function has a max, opens downwards, upside-down bowl). If "a" is positive, then the range is y >= c - (b^2)/(4a)

If "a" is negative, then the range is y <= c - (b^2)/(4a)

Answer 9

Y-intercept is easy. Just put x=0 into the equation and get your "y" value.

Answer 10

As for x-intercept(s), there may or may not be any. Again, you have 0, 1, or 2 real zeros. Determined by the discriminant in the quadratic formula (see the sqrt in upcoming formula). If what is under the sqrt is not negative, then you get at least one real zero. Also, use this formula to get your x-inteercept(s) , the real zeros, if any:
\[x = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
This is a sort-of crash-course in quadratics. Very compressed. You'll have to ask questions at this point.

Answer 11

What I gave you is the generalization of the formula and its characteristics. From this, you can treat any quadratic since it uses a, b, and c. The answers do look a little hairy, but once you do the subsitutions, they simplify. If you can train yourself to learn from formulas, you can generalize principles and see the big picture.