How do you write the quadratic in vertex form given #y=-4x^2+12x+5#?

Question

Kennedy Adams · Accepted Answer

#y=-4(x-3/2)^2+14# #"the equation of a parabola in "color(blue)"vertex form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))# #"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"# #"to obtain this form use the method of "color(blue)"completing the square"# #• " the coefficient of the "x^2" term must be 1"# #"factor out "-4# #y=-4(x^2-3x-5/4)# #• " add/subtract "(1/2"coefficient of the x-term")^2" to"# #x^2-3x# #y=-4(x^2+2(-3/2)x color(red)(+9/4)color(red)(-9/4)-5/4)# #color(white)(y)=-4(x-3/2)^2-4(-9/4-5/4)# #color(white)(y)=-4(x-3/2)^2+14larrcolor(red)"in vertex form"#

Abigail Allen · Answer

undefined To write a quadratic function in vertex form given $ y = -4x^2 + 12x + 5 $, follow these steps:

1. Complete the square on the quadratic expression.
2. Rewrite the quadratic expression in the form $ y = a(x - h)^2 + k $, where $ (h, k) $ represents the vertex of the parabola.

Given the quadratic expression $ y = -4x^2 + 12x + 5 $:

1. Factor out the coefficient of $ x^2 $ from the $ x^2 $ and $ x $ terms: $ y = -4(x^2 - 3x) + 5 $.
2. Complete the square by adding and subtracting $ (\frac{b}{2})^2 $, where $ b $ is the coefficient of the $ x $ term: $ y = -4(x^2 - 3x + (\frac{3}{2})^2 - (\frac{3}{2})^2) + 5 $.
3. Simplify the expression inside the parentheses: $ y = -4((x - \frac{3}{2})^2 - (\frac{9}{4})) + 5 $.
4. Expand the expression and combine like terms: $ y = -4(x - \frac{3}{2})^2 + 9 + 5 $.
5. Simplify: $ y = -4(x - \frac{3}{2})^2 + 14 $.

So, the quadratic function in vertex form is $ y = -4(x - \frac{3}{2})^2 + 14 $.