# SOLUTION: I am learning basically how to graph quadratic equations. I dont really understand how to find the coordinates of the vertex. I also dont understand how to write an equation of the

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: I am learning basically how to graph quadratic equations. I dont really understand how to find the coordinates of the vertex. I also dont understand how to write an equation of the      Log On

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 Question 24062: I am learning basically how to graph quadratic equations. I dont really understand how to find the coordinates of the vertex. I also dont understand how to write an equation of the axis of symmetry. Can you please help me?Answer by AnlytcPhil(1278)   (Show Source): You can put this solution on YOUR website!```I am learning basically how to graph quadratic equations. I dont really understand how to find the coordinates of the vertex. I also don't understand how to write an equation of the axis of symmetry. Can you please help me? For the equation y or f(x) = a(x-h)2 + k The vertex is this point: (h, k) The equation of the axis of symmetry is this: x = h =============================================================== For the equation y or f(x) = ax2 + bx + c The vertex is this point: ( -b/(2a), c - b2/(4a) ) The equation of the axis of symmetry is this: x = -b/(2a) ------------------------------------------------------------- Example 1: y = -4(x+4)2-3 Compare to y = a(x-h)2+k -h = +4, so h = -4 k = -3. The vertex is this point: (h, k) = (-4,-3) The equation of the axis of symmetry is this: x = h, or x = -4 ================================================================== Example 1: Find the vertex of f(x) = 2x2 - 12x + 19 Compare to f(x) = ax2 + bx + c a = 2; b = -12; c = 19 The vertex is this point: ( -b/(2a), c - b2/(4a) ) = ( -(-12)/(2·2), 19 - (-12)2/(4·2) ) ( 12/4, 19 - 144/8 ) = ( 3, 19 - 18) = (3,1) The equation of the axis of symmetry is this: x = -b/(2a) x = -(-12)/(2·2) x = 3 Edwin AnlytcPhil@aol.com```