complete square of quadratic and graph... Help please!!

[PepprRN]

Here is what I was given: f(x)=-x^2+6X+4

I am supposed to get the a) standard form of the quadratic
b) find the vertex and the x and y intercepts
c) sketch the graph
I know I have to complete the square, but the (-) in front of x^2 is giving me problems... if I factor out (-) =-(x^2-6x-4) which when I complete the square I get -(x^2-6x-9)-4-9 = -(x-3)^2-13??? The book answer is -(x-3)^2+13. my wy gives me a vertex of (-3,-13) and the answer in the book is (3,13)....what am I doing wrong with that *$#&!! negative sign in front of x^2 in the original equation????

 

[Deleted member 4993]

Here is what I was given: f(x)=-x^2+6X+4

I am supposed to get the a) standard form of the quadratic
b) find the vertex and the x and y intercepts
c) sketch the graph
I know I have to complete the square, but the (-) in front of x^2 is giving me problems... if I factor out (-) =-(x^2-6x-4) which when I complete the square I get -(x^2-6x-9)-4-9 = -(x-3)^2-13??? The book answer is -(x-3)^2+13. my wy gives me a vertex of (-3,-13) and the answer in the book is (3,13)....what am I doing wrong with that *$#&!! negative sign in front of x^2 in the original equation????

Just carry the negetive sign all the through - untill the last step.


-x² + 6x + 4

= -[x² - 6x - 4]

= - [x² - 6x + 9 - 9 - 4]

= - [(x-3)² - 13]

= - (x-3)² + 13

The sign in front of the x² tells you whether the parabola is opening up or opening down. The negetive sign means it is opening down. The vertex would be at (3,13) → at x = 3, the f(x) will reach the maximum value (vertex).

 

[srmichael]

Here is what I was given: f(x)=-x^2+6X+4

I am supposed to get the a) standard form of the quadratic
b) find the vertex and the x and y intercepts
c) sketch the graph
I know I have to complete the square, but the (-) in front of x^2 is giving me problems... if I factor out (-) =-(x^2-6x-4) which when I complete the square I get -(x^2-6x-9)-4-9 = -(x-3)^2-13??? The book answer is -(x-3)^2+13. my wy gives me a vertex of (-3,-13) and the answer in the book is (3,13)....what am I doing wrong with that *$#&!! negative sign in front of x^2 in the original equation????

When factoring out coefficient from the x^2 term, I find it easier to factor it out of only the terms with the variable x.

\(\displaystyle f(x)=-(x^2-6x)+4\)

\(\displaystyle f(x)=-(x^2-6x+9)+4+9\) ---> In this step, because of the negative sign that was factored out, we actually added a -9 when completing the square, thus we need to compensate for this by adding a +9 which leads to the +13.

\(\displaystyle f(x)=-(x-3)^2+13\)

Make sense?

 

[PepprRN]

thanks!!

see what I was doing wrong.....I think

When you are looking for the vertex, you have to account for that (-) sign in front of (x-3)²....thus giving me a vertex of (3,13) right? And that (-) means my parabola is going to point down opening up from the (3,13) vertex. Right???

BTW srmichael.....love the quote!!

 

[srmichael]

see what I was doing wrong.....I think

When you are looking for the vertex, you have to account for that (-) sign in front of (x-3)²....thus giving me a vertex of (3,13) right? And that (-) means my parabola is going to point down opening up from the (3,13) vertex. Right??? "up" should say "down"

BTW srmichael.....love the quote!!

Correct, except the parabola is opening DOWN from the (3,13) vertex, not up. You said "point down", then said "opening up" which is contradictory.

 

[PepprRN]

your right.....

I was getting into same situation with next problem, so I looked it up. The parabola y=x² in f(x)=-(x-3)²+13 the (-) means it opens downward, the -3 means it shifts to the right 3 units (making it 3) and the 13 moves it up 13 units, so vertex is (3,13)??? Do I have it??

 

[srmichael]

I was getting into same situation with next problem, so I looked it up. The parabola y=x² in f(x)=-(x-3)²+13 the (-) means it opens downward, the -3 means it shifts to the right 3 units (making it 3) and the 13 moves it up 13 units, so vertex is (3,13)??? Do I have it??

Yeeeuuuuup.

 

[PepprRN]

Yaaaaaay!!

"By jove, I think she's got it!" I feel like Eliza Doolittle!! <------(jumping up and down)

It would be much easier if they just said take y=x^2, invert it, shift it 3 to the right and move it up on the graph 13.....but to get that from
f(x)=-x²+6x+4.....

I guess if golf was easy, Tiger wouldn't be a billionaire and if this was easy there would be no need for this site!!

Thanks again, so much!!