SOLUTION: Hi, I'm having trouble with the following function. Can I please have some help solving? A quadratic function is given. f(x) = 10x^2 +x - 1 (a) Express the quadratic function

Algebra ->  Equations -> SOLUTION: Hi, I'm having trouble with the following function. Can I please have some help solving? A quadratic function is given. f(x) = 10x^2 +x - 1 (a) Express the quadratic function       Log On


   

Question 127805: Hi, I'm having trouble with the following function. Can I please have some help solving?
A quadratic function is given.
f(x) = 10x^2 +x - 1
(a) Express the quadratic function in standard form.
f(x) = a(x - h)^2 + k where
a =
h =
k =
(c) Find its minimum or maximum value.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)



y=10+x%5E2%2B1+x-1 Start with the given equation


y%2B1=10+x%5E2%2B1+x Add 1 to both sides


y%2B1=10%28x%5E2%2B%281%2F10%29x%29 Factor out the leading coefficient 10


Take half of the x coefficient 1%2F10 to get 1%2F20 (ie %281%2F2%29%281%2F10%29=1%2F20).

Now square 1%2F20 to get 1%2F400 (ie %281%2F20%29%5E2=%281%2F20%29%281%2F20%29=1%2F400)




y%2B1=10%28x%5E2%2B%281%2F10%29x%2B1%2F400-1%2F400%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F400 does not change the equation



y%2B1=10%28%28x%2B1%2F20%29%5E2-1%2F400%29 Now factor x%5E2%2B%281%2F10%29x%2B1%2F400 to get %28x%2B1%2F20%29%5E2


y%2B1=10%28x%2B1%2F20%29%5E2-10%281%2F400%29 Distribute


y%2B1=10%28x%2B1%2F20%29%5E2-1%2F40 Multiply


y=10%28x%2B1%2F20%29%5E2-1%2F40-1 Now add %2B1 to both sides to isolate y


y=10%28x%2B1%2F20%29%5E2-41%2F40 Combine like terms



Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=10, h=-1%2F20, and k=-41%2F40. 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%2F20,-41%2F40) and the parabola opens upward since a%3E0


Check:

Notice if we graph the original equation y=10x%5E2%2B1x-1 we get:

graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C10x%5E2%2B1x-1%29 Graph of y=10x%5E2%2B1x-1. Notice how the vertex is (-1%2F20,-41%2F40).


Notice if we graph the final equation y=10%28x%2B1%2F20%29%5E2-41%2F40 we get:

graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C10%28x%2B1%2F20%29%5E2-41%2F40%29 Graph of y=10%28x%2B1%2F20%29%5E2-41%2F40. Notice how the vertex is also (-1%2F20,-41%2F40).


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



c)

The min or max occurs at the vertex and the max/min value is the y value of the vertex. So in this case the minimum is -41%2F40