How do you find a vertex by completing the square? Ex. y =4x^2+20x+17

Question

How do you find a vertex by completing the square?  Ex. y =4x^2+20x+17

anonymous · Answer

To complete the squre you have to make the lead coefficient +1, so you can factor it out of the first two terms.  Then complete the squre by taking half of the 5 and squaring it.  Now we can just add 25/4 to an equation so we have to subtract an equaivalent amount on the same side of the equation.  But you have to consider that the 25/5 is, like ever term inside the parantheses, is being mutliplied by 4.  And so we compensate by subtracting 4*(25/4)=25 $$y =4x^2+20x+17=4(x^2+5x+\frac{25}{4})+17-25$$Now the expression inside the parantheses is a perfect square trinomial, and so we factor it:$$y=4(x+\frac{5}{2})^2-8$$This isthe standard form or the vertex form.  In general $$y=a(x-h)^2+k$$has vertex (h, k).  So in your problem h=-5/2 and k=-8, so your vertex is (-5/2, -8).

anonymous · Answer

Thank you!!! You've helped me understand this faster than my math teacher!!