Finding vertex of quadratic equation?

[skyza]

I have to find the vertex, axis, domain, & range of this quadratic equation:

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

I can't remember how to do the vertex. Is it y= -b/2a?

Also, I tried factoring the equation, but I think I messed up. I ended up with:

f(x)= .25x^2 +.5x-2.75



Thanks

 

[LCKurtz]

I have to find the vertex, axis, domain, & range of this quadratic equation:

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

I can't remember how to do the vertex. Is it y= -b/2a?

Also, I tried factoring the equation, but I think I messed up. I ended up with:

f(x)= .25x^2 +.5x-2.75



Thanks

I assume your equation is$$
f(x) = -\frac 1 2(x+1)^2-3$$If so, it is already in the form you need to locate the vertex. Note that the $(x+1)^2$ factor is always greater than or equal zero. Its contribution will be least when it is 0, which happens when $x=-1$. So your vertex is at $(-1,-3)$. No need to memorize formulas; just complete the square like in this problem.

 

[skyza]

Thanks for the help.

I'm trying to find the vertex of:

f(x)= -3x^2 +24x-46

and I ended up with:

-3x+144-sqrtof 190

 

[QuarkCharmer]

What are your a and b terms?

The x coordinate of the vertex is $\frac{-b}{2a}$, once you have that, you need another coordinate right?

 

[LCKurtz]

Thanks for the help.

I'm trying to find the vertex of:

f(x)= -3x^2 +24x-46

Complete the square on that. Then do what I showed you in my previous post. You do know how to complete the square, right?

 

[skyza]

Complete the square on that. Then do what I showed you in my previous post. You do know how to complete the square, right?

I tried completing the square for:

f(x)= -3x^2 +24x-46

That's how I ended up with that.
I divided 24/2 and added it to both sides and got

-3x+144-sqrtof 190

 

[LCKurtz]

I tried completing the square for:

f(x)= -3x^2 +24x-46

That's how I ended up with that.
I divided 24/2 and added it to both sides and got

-3x+144-sqrtof 190

No. You have to factor out the -3 to start, complete the square on the inside, and balance it out:
$f(x) = -3(x^2 -8x + ?) -46 + ?$

 

[skyza]

No. You have to factor out the -3 to start, complete the square on the inside, and balance it out:
$f(x) = -3(x^2 -8x + ?) -46 + ?$

Alright, I've gotten to:

-3(x-4)(x-4)=62
or
-3(x-4)^2=62
or
-3(x-4)^2 -62=0

What do I do from here?

 

[skyza]

I've figured it out.

 

[LCKurtz]

Alright, I've gotten to:

-3(x-4)(x-4)=62
or
-3(x-4)^2=62
or
-3(x-4)^2 -62=0

What do I do from here?

You don't set it equal to anything. What you should have written is$$
f(x) = -3(x^2-8x+16)-46+48 = -3(x-4)^2+2$$From there it is clear that the vertex is $(4,2)$.

 

[skyza]

You don't set it equal to anything. What you should have written is$$
f(x) = -3(x^2-8x+16)-46+48 = -3(x-4)^2+2$$From there it is clear that the vertex is $(4,2)$.

Why did you add 48 to -46 and instead of adding 16?

 

[LCKurtz]

You don't set it equal to anything. What you should have written is$$
f(x) = -3(x^2-8x+16)-46+48 = -3(x-4)^2+2$$From there it is clear that the vertex is $(4,2)$.

Why did you add 48 to -46 and instead of adding 16?

I didn't just add 16. There is a -3 out in front so I really put in -48 and I have to take it back out if I'm not to change the equation.

 

[skyza]

I didn't just add 16. There is a -3 out in front so I really put in -48 and I have to take it back out if I'm not to change the equation.

That's what I thought. Thanks!