skyza said: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
skyza said:Thanks for the help.
I'm trying to find the vertex of:
f(x)= -3x^2 +24x-46
LCKurtz said: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 said: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 said: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 said: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?
LCKurtz said: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)##.
LCKurtz said: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 said:Why did you add 48 to -46 and instead of adding 16?
LCKurtz said: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.
The vertex of a quadratic equation is the highest or lowest point on the graph of the equation. It is also known as the maximum or minimum point.
To find the vertex of a quadratic equation, you can use the formula x = -b/2a, where a and b are coefficients of the equation in the form ax^2 + bx + c. This formula will give you the x-coordinate of the vertex. To find the y-coordinate, substitute the x-coordinate back into the equation.
Yes, the vertex of a quadratic equation can be negative. The x-coordinate of the vertex can be negative if the parabola opens downwards, and the y-coordinate can be negative if the vertex is below the x-axis.
If the coefficient of the x^2 term is positive, the parabola will open upwards and have a minimum point. If the coefficient is negative, the parabola will open downwards and have a maximum point. This can also be determined by looking at the sign of the coefficient of the x^2 term in the equation.
Yes, the vertex is the only point of symmetry on a parabola. This means that if you draw a line through the vertex, the two sides of the parabola will be mirror images of each other.