What is the vertex form of the equation?
y = -x^2+12x-4
My work:
-(x^2 + 12x)-4
-(x^2 + 12x + 36)-4 + 36
Y= -(x+36)^2 + 32
Is this correct? I am not sure if I did this right or not? Thank you!
To find the vertex form of the equation y = -x^2 + 12x - 4, you need to complete the square. Here's the step-by-step process:
1. Begin with the given equation: y = -x^2 + 12x - 4.
2. Group the first two terms together: -(x^2 - 12x) - 4.
3. To complete the square, take half of the coefficient of the x-term (12/2 = 6) and square it (6^2 = 36).
4. Add and subtract the value obtained in step 3 inside the parentheses: -(x^2 - 12x + 36 - 36) - 4.
5. Rearrange the equation: -(x^2 - 12x + 36) + 36 - 4.
6. Simplify: -(x - 6)^2 + 32.
So, the correct vertex form of the equation y = -x^2 + 12x - 4 is y = -(x - 6)^2 + 32.
Your work is almost correct, you just made a small mistake when simplifying. The constant term -4 should be added to 36, not subtracted. So it should be: y = -(x+36)^2 + 32.
To find the vertex form of the equation, you need to complete the square. It seems like you are on the right track, but there is a small mistake in your calculations.
Starting from y = -(x^2 + 12x) - 4, you correctly identified the coefficient of x as 12/2 = 6. To complete the square, you need to add and subtract (6^2) = 36 inside the parentheses:
y = -(x^2 + 12x + 36 - 36) - 4
Now, you can combine like terms inside the parentheses:
y = -(x^2 + 12x + 36) + 36 - 4
Next, you can factor the perfect square trinomial inside the parentheses:
y = -((x + 6)^2 - 4) + 36
Finally, distribute the negative sign outside the parentheses and simplify:
y = -(x + 6)^2 + 4 + 36
This simplifies to:
y = -(x + 6)^2 + 40
So, the correct vertex form of the equation is y = -(x + 6)^2 + 40.
my try:
x^2 -12 x = -y - 4
x^2 - 12 x + 6^2 = -y -4 + 36
(x-6)^2 = -y + 32
so
(y-32) = -(x-6)^2
vertex at (6 , 32)