i was wondering how i would write an equation in standard form for y= -3x^2-9x-6any help is greatly appreciated! (:
You should know that you may write the quadratic equation using three forms: standard form, vertex form and factored form.
The standard form is exactly the form that the problem provides:
`y = -3x^2 - 9x - 6`
The vertex form is `y = a(x-h)^2 + k` , where (h,k) represents the vertex of parabola and you may notice that you cannot compare the vertex form to the given form since they does no look alike.
The factored form is `y = a(x-x_1)(x-x_2)` and as you may see, this form cannot be compared to the given form.
Hence, expressing the quadratic in standard form yields `y = -3x^2 - 9x - 6.`
The standard form is also the vertex form. THe best way to change it is completing of squares.
Y=-3x2-9x-6 factor the coefficient of x2 from the other terms
-3(x2+3x)-6 divide the coef. Of x2 by 2, square your result, then add and supply it into the equation i.e. (3/2)2 =9/4
-3(x2+3x+9/4-9/4)-6 bring the last term inside the bracket outside the bracket
-3(x2+3x+9/4)-3(-9/4)-6 factor the perfect square inside the bracket then simplify i.e.
The vertex is (-3/2, +3/4)
To prove this answer, use the graphing calculator.(sorry, this answer may contradict what others have done but this is the right answer.)