I need help in finding the vertex of a parabola using an equation in this form: Ax^2 + Bx+C
I have a test on this tomorrow.
For example: 3x^2 +1x - 2
How would I find the vertex in this?
Unless you are instructed to do it in a certain way, here is the simplest and quickest way:
For any parabola in the form
y + Ax^2 + Bx + C
the x of the vertex is -B/(2A)
so in our case -1/3
You then sub that back into your equation to find the y of the vertex,
for ours ...
y = 3(1/9) + 1/3 - 2 = 2/3 - 2
= -4/3
Let me know if this answers is sufficient.
Thanks!
To find the vertex of a parabola in the quadratic equation form, Ax^2 + Bx + C, you can use the formula:
x = -B / 2A
Let's apply this formula to the given quadratic equation: 3x^2 + 1x - 2.
In this case, A = 3, B = 1, and C = -2.
First, calculate the x-coordinate of the vertex using the formula:
x = -B / 2A
= -(1) / 2(3)
= -1 / 6
Now, substitute this x-coordinate value back into the equation to find the y-coordinate of the vertex:
y = 3(-1/6)^2 + 1(-1/6) - 2
= 3(1/36) - 1/6 - 2
= 1/12 - 1/6 - 2
= 1/12 - 2/12 - 24/12
= -25/12
Therefore, the vertex of the parabola defined by the equation 3x^2 + 1x - 2 is (-1/6, -25/12).