Question

Am I right or wrong?

f(x)=-(x+1)^2+4
vertex: (-1,4)
x= .4
x= -2.4
y= 5

Answer 1

@Hero

Answer 2

I agree with your vertex. What does the rest of the information represent. I assume the rest of what you posted are the x and y intercepts.

Answer 3

Yes they're. But, I'm not entirely positive I got them properly, I can type up the equation, if you'd like.

Answer 4

\(\bf y = -(x+1)^2+4\implies 0=-x^2-2x-1+4\implies 0=-x^2-2x+3\\ \quad \\
\textit{using the quadratic formula}\\ \quad \\
x= \cfrac{ - (-2) \pm \sqrt { (-2)^2 -4(-1)(3)}}{2(-1)}\implies x= \cfrac{ 2 \pm \sqrt { 4+12}}{-2}\)

Answer 5

Here's the graph of it:

https://www.desmos.com/calculator/lg1z24reec

Answer 6

pellet, it was that formula?! I was using the (a+h)^2+k

Answer 7

There's no reason to use the quadratic formula to find the x and y intercepts.

Answer 8

I'll fix it up now. Dx

Answer 9

Wait, I don't understand... Which formula then?

Answer 10

Instead, what you do is:
First remember y = f(x)
To find the x intercept, set y = 0, then solve for x:
y=-(x+1)^2+4
0 = -(x+1)^2 + 4
(x+1)^2 = 4
x + 1 = ±2
x = ±2 - 1
`x = 1`
`x = -3`

To find the y-intercept, set x = 0, then solve for y
y = -(0+ 1)^2 + 4
y = -(1)^2 + 4
y = -1 + 4
`y = 3`

Answer 11

I see where I messed up now. Thank you!

Answer 12

Okay, let me make sure that I'm absolutely doing it right.

f(x)=(x+4)^2-2
vertex= (-4,-2)

0=(x+4)^2-2
2=(x+4)^2
1.4= x+4
-1.4-4 = -5.4
1.4-4=-2.6

f(0)=(0+4)^2-2
4^2-2
16-2
14

Is that done right?''
@Hero

Answer 13

Correct

Answer 14

Yay!!

Okay, now these two are different but I'm positive they're right.

f(x)=-x^2+2x+3
-(2/2(-1)) = -2/-2 = 1

0=-x^2+2x+3
0=(x-1)(x+3)
x=1 x=-3

y=c so... y=3

f(1)=1^2+2+3
1+2+3
y=5

(1,5)

Sorry for being bothersome, by the way;;

Answer 15

and then there's


f(x)=2x^2-4x-6
-(-4/2(2)= 4/4 = 1

f(1)= 2^2-4-6
4-4-6
0-6
-6

(1,-6)

f(0)=2(0)^2-4(0)-6
0-0-6
y=-6

0=2x^2-4x-6
2(x^2-2x-3)
2(x+1)(x-3)

x= 1.4
x= -1
x= 3

Answer 16

@Hero

Answer 17

I'm about to go take a nap. ttyl

Answer 18

@phi , do you think you can help me? :c

Answer 19

***
0=-x^2+2x+3
0=(x-1)(x+3)
x=1 x=-3
***

the minus out front changes things. I would factor it out to get
-( x^2 -2x -3) = 0
(x-3)(x+1)=0
x=3 , x= -1

Answer 20

Yeah--- I see where I got it wrong... But other than that, the rest is fine, right? c:

Answer 21

***
f(x)=2x^2-4x-6
-(-4/2(2)= 4/4 = 1

f(1)= 2^2-4-6
***
x value of the vertex is 1.
the y value is 2*1^2 - 4*1 -6 = 2 - 4 - 6= 2-10= -8

Answer 22

it looks like you have the right idea, but you have to be more careful when doing the work.

Answer 23

Ahh-- Dang, I didn't realize that. But thank you so much!! I really appreciate it.

Answer 24

Would you mind helping me with four other equations?

Answer 25

@TheHero, when doing these, you can easily check your own work.