What is the vertex form of the equation? y = -x^2 + 12x - 4. I have no clue how to do this.
Thanks for helping them, but remember, the education system is flawed. Not all teachers do their job (the place I teach at, the other teachers just don't care), not all students understand the material given (some students coming from another school, or had a different learning pace in a previous grade, might not have done the material at all before the new teacher they have expects them to have already learned it), not all students learn the same way (some might not solve word problems as well, some might be visual learners, some might need things read to them, some might need a hands on approach, everyone is different and school only teaches you one way, and if you don't fit the mold, you get left behind in favor of the ones that do), and not all students respect that others may not get it as fast as they do (students who understand it faster get bored easy, and tend to disrupt the learning for the ones who don't get it as quickly).
To convert the equation y = -x^2 + 12x - 4 into vertex form, follow these steps:
Step 1: Identify the coefficients a, b, and c in the given equation.
In this case, a = -1, b = 12, and c = -4.
Step 2: Find the x-coordinate of the vertex.
The formula for finding the x-coordinate of the vertex in terms of a and b is given by x = -b / (2a). Substitute the values of a and b into this formula.
x = -12 / (2*(-1))
x = -12 / (-2)
x = 6
Step 3: Substitute the x-coordinate of the vertex (6) back into the original equation to find the y-coordinate.
y = -(6)^2 + 12(6) - 4
y = -36 + 72 - 4
y = 32
So, the vertex has the coordinates (6, 32).
Step 4: Rewrite the equation using the vertex form of a quadratic equation, which is y = a(x - h)^2 + k, where (h, k) are the coordinates of the vertex.
Substituting the vertex coordinates into the vertex form, we get:
y = -1(x - 6)^2 + 32
Therefore, the vertex form of the equation y = -x^2 + 12x - 4 is y = -1(x - 6)^2 + 32.
To find the vertex form of a quadratic equation, you need to rewrite the equation in the form y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola. The process involves completing the square. Let's go step by step to find the vertex form of the given equation, y = -x^2 + 12x - 4:
Step 1: Rearrange the equation so that the quadratic term (x^2) is first, the linear term (x) is second, and the constant term (-4) is last:
y = -x^2 + 12x - 4
Step 2: Group the x-terms together:
y = -(x^2 - 12x) - 4
Step 3: Complete the square inside the parentheses:
To complete the square, take half of the coefficient of the x-term (-12), square it, and add it inside the parentheses and outside the parentheses (since you subtracted it):
y = -(x^2 - 12x + (-12/2)^2) + (12/2)^2 - 4
= -(x^2 - 12x + 36) + 36 - 4
Step 4: Simplify:
y = -(x^2 - 12x + 36) + 32
Step 5: Rewrite the equation in the vertex form:
y = - (x - 6)^2 + 32
So, the vertex form of the equation y = -x^2 + 12x - 4 is y = - (x - 6)^2 + 32. The vertex of the parabola is (6, 32).
If you have no clue, then clearly you have not studied your material, since it must have been covered.
If a parabola has vertex at (h,k), then the equation is
y = a(x-h)^2+k
You have
y = -x^2+12x-4
= -(x^2-12x) -4
Now recall that (x+n)^2 = x^2+2nx+n^2
That means that you have 2n = -12, or n=-6.
So, add (-6)^2=36 inside the parentheses and you have a perfect square. But, having done that, you have changed the expression, so you have to subtract it as well, so the value is unchanged:
y = -(x^2-12x+36) - 4 + 36
y = -(x-6)^2 + 32
To check you answer, expand things out; it should give you what you started with:
-(x-6)^2+32
= -(x^2-12x+36)+32
= -x^2+12x-36+32
= -x^2+12x-4