SOLUTION: Hi Can I have some help concerning functions? A quadratic function is given. f(x) = 6x^2 + x + 1 (a) Express the quadratic function in standard form. f(x) = a(x - h)^2 + k

Algebra ->  Equations -> SOLUTION: Hi Can I have some help concerning functions? A quadratic function is given. f(x) = 6x^2 + x + 1 (a) Express the quadratic function in standard form. f(x) = a(x - h)^2 + k       Log On


   

Question 127713: Hi Can I have some help concerning functions?
A quadratic function is given.
f(x) = 6x^2 + x + 1
(a) Express the quadratic function in standard form.
f(x) = a(x - h)^2 + k where
a =
h =
k =
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

y=6+x%5E2%2B1+x%2B1 Start with the given equation


y-1=6+x%5E2%2B1+x Subtract 1 from both sides


y-1=6%28x%5E2%2B%281%2F6%29x%29 Factor out the leading coefficient 6


Take half of the x coefficient 1%2F6 to get 1%2F12 (ie %281%2F2%29%281%2F6%29=1%2F12).

Now square 1%2F12 to get 1%2F144 (ie %281%2F12%29%5E2=%281%2F12%29%281%2F12%29=1%2F144)




y-1=6%28x%5E2%2B%281%2F6%29x%2B1%2F144-1%2F144%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F144 does not change the equation



y-1=6%28%28x%2B1%2F12%29%5E2-1%2F144%29 Now factor x%5E2%2B%281%2F6%29x%2B1%2F144 to get %28x%2B1%2F12%29%5E2


y-1=6%28x%2B1%2F12%29%5E2-6%281%2F144%29 Distribute


y-1=6%28x%2B1%2F12%29%5E2-1%2F24 Multiply


y=6%28x%2B1%2F12%29%5E2-1%2F24%2B1 Now add 1 to both sides to isolate y


y=6%28x%2B1%2F12%29%5E2%2B23%2F24 Combine like terms



Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=6, h=-1%2F12, and k=23%2F24. Remember (h,k) is the vertex and "a" is the stretch/compression factor. Also "a" tells us which direction the parabola opens.



So in this case the vertex is (-1%2F12,23%2F24) and the parabola opens upward since a%3E0


Check:

Notice if we graph the original equation y=6x%5E2%2B1x%2B1 we get:

graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C6x%5E2%2B1x%2B1%29 Graph of y=6x%5E2%2B1x%2B1. Notice how the vertex is (-1%2F12,23%2F24).


Notice if we graph the final equation y=6%28x%2B1%2F12%29%5E2%2B23%2F24 we get:

graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C6%28x%2B1%2F12%29%5E2%2B23%2F24%29 Graph of y=6%28x%2B1%2F12%29%5E2%2B23%2F24. Notice how the vertex is also (-1%2F12,23%2F24).


So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.