I need to write this parabola question 2(x+3)^2 -1 in general form. Not sure how to do it. Do I sqaure whats in brackets first? x^2 + 6x +9 then *by 2 so 2x +12x +18 - 1 = 0 Would this be the answer or am I doing it wrong? Any help appreciated.
yes, you are on the right track
y = 2(x+3)^2 -1
= 2(x^2 + 6x + 9) - 1
= 2x^2 + 12x + 18 - 1
= 2x^2 + 12x + 17
whats the
area of a sqaure in sqaure metre?
help me !!
To write the given parabola equation, 2(x+3)^2 - 1, in general form, you need to expand the square and simplify the expression. Here's how you can do it step by step:
1. Start by expanding the square:
2(x+3)^2 = 2(x+3)(x+3)
Using the FOIL method to multiply, you get:
= 2(x^2 + 3x + 3x + 9)
= 2(x^2 + 6x + 9)
2. Next, distribute the coefficient 2 to each term inside the parentheses:
= 2x^2 + 12x + 18
3. Finally, subtract 1 from the expression to obtain the general form:
2x^2 + 12x + 18 - 1 = 0
Simplifying the equation gives you the final answer:
2x^2 + 12x + 17 = 0
Therefore, the general form of the given parabola equation is 2x^2 + 12x + 17 = 0.