# SOLUTION: I need help solving this the vertex of the graph of this function: f(x)=-4x^2+24x-40 I also need the axis of the graph of function for: f(x)=4x^2-32x+27 As well as the quad

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: I need help solving this the vertex of the graph of this function: f(x)=-4x^2+24x-40 I also need the axis of the graph of function for: f(x)=4x^2-32x+27 As well as the quad      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Question 533431: I need help solving this the vertex of the graph of this function: f(x)=-4x^2+24x-40 I also need the axis of the graph of function for: f(x)=4x^2-32x+27 As well as the quadratic function in vertex form of: y=x^2+8x Thanks for any helpAnswer by lwsshak3(6505)   (Show Source): You can put this solution on YOUR website!Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex .. I need help solving this the vertex of the graph of this function: f(x)=-4x^2+24x-40 complete the square: y=-4(x^2-6x+9)-40+36 y=-4(x-3)^2-4 vertex: (3,-4) .. I also need the axis of the graph of function for: f(x)=4x^2-32x+27 complete the square y=4(x^2-8x+16)+27-64 y=4(x-4)^2-37 vertex: (4,-37) axis of symmetry: x=4 (parabola opens upwards) .. As well as the quadratic function in vertex form of: y=x^2+8x complete the square y=(x^2+8x+16)-16 y=(x+4)^2-16 vertex: (-4,-16)