Equation: y=3x^2+6x+4 First you have to find the y-intercept. To find the y-intercept you always look at the number without an x attached to it. So in this example the “4” would be the y intercept. (0,4)
Second: The second thing you have to do is find the vertex. To find the vertex, you have to set up an equation . Y= -3(X^2-2x+?)+4-?*-3 We left the two numbers as a question mark because we have to find them. The first blank is always left positive, while the other is always left as negative. Next you take the number in parentheses, multiply it by half and then square it.
Cont. So our number with the x would be -2. -2*1/2=-1^2=-1 This gives us our “Axis of symmetry” Also it gives us the number that fills both of our blanks which is one.
Third: Next after you have found the remaining blanks you work the equation out. This gives you y=-3(x-1)^2=7. The seven gives us the other part of our vertex, which would make our vertex (1,7)
Fourth: The fourth step is to find the x-intercepts. To find this we have to do the quatdratic formula. Y=-b+/ √ (-b)^2-4ac 2a Next you have to know what you’re a, b, and c are. To find them you have to go back to the original equation. Which is y=3x^2+6x+4.
Cont. So knowing that, the a comes from the number with x^2. B comes from the number with just the x, and c comes from the number by itself. So if you filled in the equation with the correct numbers it would read –(6)+/- √ (6)^2-4(-3)(4) 2(-3) -6+ √ 84= -.53-6- √ 84=2.53 -6 -6 This would give you your x-intercepts of (-.53,0) and (2.53,0)
Last: The last thing to do would be graphing your points and creating your parabola.