Equation in
vertex formπ = πππ + 4 find the Coordinates of
vertex
To convert the equation π¦ = 3π₯^2 + 4 into vertex form, we need to complete the square.
Firstly, we factor the coefficient of x^2 out of the equation, which gives us:
π¦ = 3(x^2 + 4/3)
Next, we need to add and subtract a constant term to the parenthetical expression inside the parentheses. We can do this by taking half of the coefficient of x (which is 0 in this case) and squaring it.
(0/2)^2 = 0
So we add and subtract 0 inside the parentheses:
π¦ = 3(x^2 + 4/3 + 0 - 0)
Now we can write this expression in vertex form by factoring the perfect square inside the parentheses and applying the distributive property:
π¦ = 3((x + 0)^2 + 4/3 - 0)
Simplifying this expression gives us:
π¦ = 3(x + 0)^2 + 4
Therefore, the vertex of this equation is (0,4).
To find the coordinates of the vertex of the equation in vertex form, we first need to identify the values of "h" and "k" in the equation.
The general equation in vertex form is given by: π¦ = π(π₯ β β)^2 + π
In this case, the equation is π¦ = 3π₯^2 + 4.
Comparing this with the general equation, we can see that "a" is 3, and there is no value present in the equation for "h." Therefore, we have π = 3 and β = 0.
To find "k," we substitute the values of "a," "h," and "k" into the equation. Let's substitute the values:
π¦ = 3π₯^2 + 4
Since β = 0, the equation becomes:
π¦ = 3(π₯ β 0)^2 + π
Simplifying further:
π¦ = 3π₯^2 + π
Since the equation does not have a constant term, that is the term independent of "x," we can conclude that π = 0.
Therefore, the coordinates of the vertex are (h, k) = (0, 0).