help please I am studying for a math test tomorrow and have run into a problem please if there is a math genie

Question

that can help me then please do:)

rewrite the quadratic equation function in the vertex-graphing form y=a(x-p)^+q

now rewrite the equation please show me all the work please
y=x^2-4x-5

Answer 1

Hi,

y=x^2 - 4x - 5

Complete the square on x^2 - 4x. Half of -4 is -2. (-2)^2 equals 4. Add 4 to complete the square and also subtract 4 on the end to balance that out.

y=x^2 - 4x + 4 - 5 - 4

Factor the first 3 terms on the right. Combine the last 2 terms.

y=x^2 - 4x + 4 - 5 - 4
y= (x^2 - 4x + 4) - 9
y= (x - 2)^2 - 9
This is in the form y=a(x-p)^+q where the vertex is at (2,-9).

I hope that helps!! :-)

Answer 2

Alright, so the first thing you have to know is how to expand
(x-p)^2

Expanding this gives you x^2 - 2px + p^2.

your equation is y = x^2 - 4x - 5.

if you let p = 2, then (x-2)^2 = x^2 - 4x + 4

your equation is x^2 - 4x + 4, which is very close to the equation you need.
However, you can perform some tricks on your equation, If you add and subtract 4, then you will not be changing your equation at all because you will just be adding 0 to it.

y = x^2 - 4x - 5 + 4 - 4 and now rearrange:
y = (x^2 - 4x + 4) -5 - 4

notice how the bracket stuff is the same as (x-p)^2 if p=2? So now you can rewrite that as (x-2)^2
y = (x-2)^2 - 9

So there you go, it is in the vertex graphing form.



This method is pretty useful, but you have to know how to choose the p. You can ALWAYS do this to quadratic equations. In this case, the a was 1, but if the a isn't 1, then it causes some more work to be done. I will give you an example:

y = 2x^2 + 16x - 6

In this case, the first thing you have to do is factor out a 2 from the equation:
y = 2(x^2 + 8x - 3)

Now, you do the same steps for the bracket terms. The p you choose is always going to be half of the term in front of the x (in this case, the term is 8 so then,p=4)
y = 2[(x^2 + 8x + 16) - 16 - 3]

The 16 that you add and subtract is just p^2.

y = 2[(x+4)^2 - 19]

Now, here is a tricky part. Your solution is almost in the correct form, except you have to take that 19 out of the brackets. The way you would do this is to multiply 2 by everything in the square brackets (don't expand the round brackets).

y = 2*(x+4)^2 - 2*19
y = 2(x+4)^2 - 38

There is the final vertex graphing form solution.

I hope you understood that. Good luck on your test. Laterz

Answer 3

You need to use the complete the square method to rewrite this equation.

y=x^2-4x-5

y=(x^2-4x+c)-5-c

Where c is (1/2b)^2 (In this case b is -4 so c ends up being (-2)^2 or 4). Note that you have to subtract c to cancel out you adding c.

y=(x^2-4x+4)-5-4

y=(x-2)^2-9

which is in the form u were looking for.

Answer 4

The vertex form of a parabola with vertex (p,q) is:

y = a(x - p)² + q

To get that form complete the square.

y = x² - 4x - 5 = (x² - 4x + 4) - 5 - 4
y = (x - 2)² - 9

The vertex of the parabola (p,q) = (2, -9)

Answer 5

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