Question

**GIVING MEDAL**


(see attachment)
Which equation best represents the graph shown above?

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

Answer 1

@JjJustin
@lizzylou169

Answer 2

Well, if you don't care why, then the graph is of
\[ f(x) = x^2 -2x + 1 = (x - 1)^2 \]
If you do care why, then perhaps.. I could try to explain =S hehe

Answer 3

Please explain! I'd really like to get this down for future assignments! :D

Answer 4

Hmm. i'll try. Not that I'm math expert myself or anything.
Sorry it's so long... I'm too tired to find a better way writing it..

Well when it comes to quadratic equations we have:
\[ f(x) = ax^2 + bx + c \]
We know that the ax^2 part doesn't change the parabola's vertex location, only the parabola's shape.

In order to move the parabola around, you need a combination of bx and c. Why?
The c is easiest to understand. It just adds a number too alllll the values in the parabola and shifts it up.

The bx is abit more tricky but not really hard to get the concept either.
Imagine the following:
\[ f(x) = x^2 - 6x \]
Let's see on what points the parabola crosses the X axis ( f(x) = 0 )
Let's write it this way:
\[
y = x^2 - 6x \\
y = 0 \\
x^2 -6x = 0 \\
x^2 = 6x
\]
It's easy to see that if x=0 this equation is right, so that's one solution. Let's find the other:
\[
x \ne 0 \\
x^2 = 6x \]
Now to divide by x (we can cuz it's not 0)
\[
x = 6
\]
That means the parabola crosses X axis at x = 0 and x = 6, so by the shape of parabola the vertex is inbetween (x = 3)
The value of the vertex now is
\[ f(3) = x^2 - 6x = 3^2 - 6 \cdot 3 = 3^2 - 2 \cdot 3 \cdot 3 = \\
=
3^2 - 2 \cdot 3^2 = -(3^2) = -9 \]
So now the vertex is in (3, -9). If you would like the vertex in (3,0) for example, you would change the equation to this:
\[ f(x) = x^2 -6x + 9 \]
to shift up all values up by 9 (including vertex ofc)

Ofc, that for practical calculations of the cross points of X axis you shoudl use the quadratic formula. I decided to include the whole way so you would see there is no special 'magic' in here..
\[
f(x) = ax^2 + bx + c \\
ax^2 + bx + c = 0 \\
ax^2 + bx = - c \\
ax^2 + bx = - c \text{ //}\cdot 4a\\
4a^2x^2 + 4abx = - 4ac \text{ //}+ b^2 \\
4a^2x^2 + 4abx + bx^2= b^2 - 4ac \\
(2ax + b)^2 = b^2 - 4ac \text{ //} \sqrt{}\\
2ax + b = \pm \sqrt{b^2 - 4ac} \\
2ax = -b \pm \sqrt{b^2 -4ac} \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Answer 5

Thank you so much!

Answer 6

Eh you didn't kill yourself? good.. thought you might hehe

Answer 7

Nah :) I just needed some help and this is pretty clear (though, i just skimmed for now). I really appreciate it :D

Answer 8

sure np =]