Write the quadratic function f(x) = x2 + 8x + 3 in vertex form?
A) f(x) = (x - 4)2 - 13
B) f(x) = (x - 4)2 + 3
C) f(x) = (x + 4)2 + 3
D) f(x) = (x + 4)2 - 13

Question

Write the quadratic function f(x) = x2 + 8x + 3 in vertex form?
A) f(x) = (x - 4)2 - 13 
B) f(x) = (x - 4)2 + 3 
C) f(x) = (x + 4)2 + 3 
D) f(x) = (x + 4)2 - 13

Genesis Allen · Accepted Answer

undefined D) f(x) = (x + 4)² - 13

Kennedy Adams · Answer

#"D": f(x)=(x+4)^2-13#

You are requested to convert the following function to vertex form:

#f(x)=x^2+8x+3#

The suggested potential fixes are:

#"A") f(x)=(x-4)^2-13# #"B") f(x)=(x-4)^2+3# #"C") f(x)=(x+4)^2+3# #"D") f(x)=(x+4)^2-13#

Converting to Vertex Form #1#. Start by placing brackets around the first two terms.

#f(x)=x^2+8x+3#

#f(x)=(x^2+8x)+3#

#2#. In order to make the bracketed terms a perfect square trinomial, we must add a "#color(darkorange)c#" term as in #ax^2+bx+color(darkorange)c#. Since #color(darkorange)c#, in a perfect square trinomial is denoted by the formula #color(darkorange)c=(color(blue)b/2)^2#, take the value of #color(blue)b# to find the value of #color(darkorange)c#.

#f(x)=(x^2+color(blue)8x+(color(blue)8/2)^2)+3#

#3#. However, adding #(8/2)^2# would change the value of the equation. Thus, subtract #(8/2)^2# from the #(8/2)^2# you just added.

#f(x)=(x^2+8x+(8/2)^2-(8/2)^2)+3#

#4#. Multiply #(-(8/2)^2)# by the #color(violet)a# term as in #color(violet)ax^2+bx+c# to bring it outside the brackets.

#f(x)=(color(violet)1x^2+8x+(8/2)^2)+3-((8/2)^2xxcolor(violet)1)#

#5#. Simplify.

#f(x)=(x^2+8x+16)+3-16#

#f(x)=(x^2+8x+16)-13#

#6#. Lastly, factor the perfect square trinomial.

#color(green)(|bar(ul(color(white)(a/a)f(x)=(x+4)^2-13color(white)(a/a)|)))#

Write the quadratic function f(x) = x2 + 8x + 3 in vertex form? A) f(x) = (x - 4)2 - 13 B) f(x) = (x - 4)2 + 3 C) f(x) = (x + 4)2 + 3 D) f(x) = (x + 4)2 - 13