How do you write #f(x)= -2x^2 + 16x +4# in vertex form?

3 Answers
Mar 26, 2016

#f(x)=-2(x-4)^2+36# with vertex at #(4,36)#

Explanation:

General vertex form is
#color(white)("XXX")f(x)=color(green)(m)(x-color(red)(a))^2+color(blue)(b)# with the vertex at #(color(red)(a),color(blue)(b))#

Given
#color(white)("XXX")f(x)=-2x^2+16x+4#

Extract the #color(green)m# component:
#color(white)("XXX")f(x)=color(green)(-2)(x^2-8x)+4#

Complete the square:
#color(white)("XXX")f(x)=color(green)(-2)(x^2-8x+color(cyan)(16))+4-color(cyan)((-2*)(16))#

Re-write as squared binomial and simplify to get vertex form
#color(white)("XXX")f(x)=color(green)(-2)(x-color(red)(4))^2+color(blue)(36)#
graph{-2x^2+16x+4 [-5.92, 26.13, 22.5, 38.54]}

The vertex form is #(x-4)^2=-1/2*(y-36)#

Explanation:

We start from the given #f(x)=-2x^2+16x+4#

Let #y=-2x^2+16x+4#

Start by factoring out the -2 from the first two terms

#y=-2(x^2-8x)+4#

We now use the -8. Divide this number by 2 then the result be squared so that we will have #(-8/2)^2=+16#

This 16 will be added and subtracted inside the grouping symbol.

#y=-2(x^2-8x)+4#

#y=-2(x^2-8x+16-16)+4#

We now have a PFT-Perfect Square Trinomial #(x^2-8x+16)=(x-4)^2#

So that we have

#y=-2(x^2-8x+16-16)+4#

#y=-2((x-4)^2-16)+4#

Put the -2 back

#y=-2(x-4)^2+32+4#

Simplify

# y=-2(x-4)^2+36#
transpose the 36 to the left of the equation

# y-36=-2(x-4)^2#

divide by -2

#(x-4)^2=-1/2*(y-36)#

God bless....I hope the explanation is useful.

Mar 26, 2016

#y = -2(x -4)^2 + 36#

Explanation:

There is another way of finding the vertex form.
x-coordinate of vertex:
#x = -b/(2a) = -16/-4 = 4#
y-coordinate of vertex:
y(4) = -32 + 64 + 4 = 36
Vertex form:
#y = a(x - 4)^2 + 36 = -2(x - 4)^2 + 36#