Transformations from Parent Functions

[KristyKat]

We got some summer homework to do before school starts in August, I'm going into A.P. Calculus and my Pre-Calculus teacher wasn't in school much to teach us...

The instructions are, "Sketch the graph of each function as accurately as possible. State the x-intercept(s), y-intercept, vertex, asymptote(s), and/or any other important point(s)."

Question 1 =
y = -(x+2)^2+3

I used a table and got:
(-4,-1)
(-3,2)
(-2,3)
(-1,2)
(0,-1)

And the vertex is (-2,3).

I don't even know where to go from here.
Is 2 the y-intercept?

 

[Aladdin]

To find the y-intercept : for x=0 --- > y =-1

To find the x-intercepts : for y =0 ---> -x^2 - 4x - 1 = 0 Quadrtic formula, their are two solutions

No verticle nor horizantal and oblique assymptote in this function .because :

when, x tends to -+infinity --- > y tends to - infinity

 

[KristyKat]

Aladdin::

Thank you so much. Okay, I understand how you got the y-intercept and I think I'm okay with the asymptotes. But I wanted to make sure I'm getting the right x-intercepts.

This is what I have so far...

x = 4 +/- (SR)-4^2-4(-1)(-1) / 2(-1)

x = 4 +/- (SR)32 / -2

x = 4+/- 4(SR)2 / -2

I think it can be simplified more but I'm not sure...

 

[Aladdin]

Aladdin::

Thank you so much. Okay, I understand how you got the y-intercept and I think I'm okay with the asymptotes. But I wanted to make sure I'm getting the right x-intercepts.

This is what I have so far...

x = 4 +/- (SR)-4^2-4(-1)(-1) / 2(-1)

x = 4 +/- (SR)32 / -2

x = 4+/- 4(SR)2 / -2

I think it can be simplified more but I'm not sure...

- x² - 4x -1 = 0

Delta = b'² - ac

Delta = (-2)² -(-1)(-1)

Delta = 4 - 1 = 3

x'=x''= (- b' -+?delta)/a

x'= (-(-2) - ?3)/-1 --- > x'=(2-?3)/-1

x''= (2 + ?3)/-1

Sorry if I wasn't clear :?

 

[KristyKat]

That's okay. =]

The thing I don't get though is why are you using a different quadratic equation than me?

 

[Aladdin]

That's okay. =]

The thing I don't get though is why are you using a different quadratic equation than me?

Yep, you're right . It depends on the country or place you're learning....

 

[KristyKat]

Oh, seriously?! I had no clue... Thanks for all of your help though!

 

[BigGlenntheHeavy]

\(\displaystyle Start \ with \ f(x) \ = \ -x^{2}, \ then \ g(x) \ = \ -(x+2)^{2}, \ then \ h(x) \ = \ -(x+2)^{2}+3.\)-[attachment=0:25z4t1b0]triple.gif[/attachment:25z4t1b0]

With g(x), shift 2 units to the left, h(x) = g(x) plus shift 3 units upward. Hence vertex = (-2,3), x intercepts (h(x) = 0), 0= -(x+2)^2+3, (x+2)^2 = 3, x+2 = ±?3, x = -2 ±?3, y intercept (x=0), y = -1, no asymptotes, domain (-?,?), range
(-?,3), axis of symmetry, x = -2.

Note: A trick. If you have a quadratic equation, say x^2-6x+4 and you want to graph it, complete the square and you get (x-3)^2-5. then your graph of x^2 is shifted 3 units to the right and 5 units downward.

 

[KristyKat]

Thanks so much. After graphing it on my calculator I was kind of confused on how it got that answer but that helped a lot, the explanation and the graph. Thanks again!