Express the quadratic function y=-2/3x2-12x-11 in standard form?
follow the same method I showed you in the previous post.
To express the quadratic function in standard form, we need to rewrite it in the form of y = ax^2 + bx + c, where a, b, and c are constants.
Given the function y = -2/3x^2 - 12x - 11, we can start by multiplying through by 3 to eliminate the fraction:
3y = -2x^2 - 36x - 33
Next, let's rearrange the terms by grouping the x^2 and x terms together:
3y = -2x^2 - 36x - 33
3y = (-2x^2 - 36x) - 33
Now, we can factor out a common factor of -2 from the first two terms:
3y = -2(x^2 + 18x) - 33
To complete the square and convert the quadratic expression inside the parentheses into a perfect square trinomial, we need to add and subtract the square of half the coefficient of x. In this case, the coefficient of x is 18, so we have:
3y = -2(x^2 + 18x + 81 - 81) - 33
Now, we can simplify the expression inside the parentheses:
3y = -2[(x + 9)^2 - 81] - 33
3y = -2(x + 9)^2 + 162 - 33
3y = -2(x + 9)^2 + 129
Finally, dividing through by 3 gives us the quadratic function in standard form:
y = (-2/3)(x + 9)^2 + 43
Therefore, the quadratic function y = -2/3x^2 - 12x - 11 in standard form is y = (-2/3)(x + 9)^2 + 43.