# SOLUTION: For homework I have to write in standard form, identify vertex, axis of symmetry, focus, and direction of opening, then graph also and I am lost! My problem is: y=-x(squared

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: For homework I have to write in standard form, identify vertex, axis of symmetry, focus, and direction of opening, then graph also and I am lost! My problem is: y=-x(squared      Log On

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 407351: For homework I have to write in standard form, identify vertex, axis of symmetry, focus, and direction of opening, then graph also and I am lost! My problem is: y=-x(squared)-12x=20 Please help me. Thank you Josie Found 2 solutions by stanbon, katealdridge:Answer by stanbon(57410)   (Show Source): You can put this solution on YOUR website!write in standard form, identify vertex, axis of symmetry, focus, and direction of opening, then graph: y = -x^2-12x=20 Comment: You have to many equal signs. I'll just assume you mean y = -x^2-12x+20 ----- Complete the square with the "x-terms". --- y-20 = -x^2-12x y-20-36 = -(x^2+12x+36) y-56 = -(x+6)^2 y = -(x+6)^2+56 h = -6; k = 56 ; 4p = -1, so p = -1/4 ---- Vertex: (-6,56) ; axis: x = -6 ; focus: (-6,55 3/4) ; opening down ======================================================================== ----- Cheers, Stan H. ========== Answer by katealdridge(100)   (Show Source): You can put this solution on YOUR website! Standard Form requires you to have all terms on one side of the = Subtract 20 from both sides. This is the polynomial in standard form. , the number in front of , , the number in front of x, and , the number with no x next to it. To find the vertex, This is the x coordinate of the vertex and the axis of symmetry. Then substitute 6 in for x in the equation: This is the y coordinate of the vertex. Vertex: (6,-56), Axis of symmetry: x=6. Direction of opening is up. That is based on a, the coefficient of the first term. If the first term is positive, then the parabola opens up, if it's negative then it opens down. The focus is found by changing the standard form equation into the focus-directrix form. <- this, by the way is not a proper way to write an equation you need to complete the square here. Take half of the second term's coefficient, -12/2 = -6. Then square that number = 36. This is the number you put in the blank spot. You need to subtract 36 at the end to make up for the adding of 36 in the polynomial. Otherwise it wouldn't be equivalent or a legal step. This is the polynomial written in a binomial squared form. Then add 56 to both sides This is now in focus-directrix form. (4p(y-k)=(x-h)^2), where (h,k) is the vertex. And the distance from the vertex to the focus is p. Since the graph opens up the focus is vertically 1/4 away from the vertex, putting it at {6,-55.75) The focus is always inside the parabola rather than outside it. That's why it's at -55.75 and not -56.25. Then you have to graph it. That part's up to you!