Gabylovesyou
  • Gabylovesyou
convert f(x) = 2x^2 - 4x + 5 into general form.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Gabylovesyou
  • Gabylovesyou
@freckles
Gabylovesyou
  • Gabylovesyou
using these steps: 1. Isolate the terms containing the x variable. 2.Complete the square and balance the equation. Factor the numerical GCF to ensure the leading coefficient of the binomial is 1. Divide the coefficient of x by 2 and square the result to create a perfect square trinomial. To keep the equation balanced, the constant added to one side of the equation must also be added to the other side of the equation. 3. Simplify. Combine like terms. Factor the perfect square trinomial. Isolate the y variable.
freckles
  • freckles
sounds like general form is vertex form

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Gabylovesyou
  • Gabylovesyou
yes.. :)
freckles
  • freckles
\[f(x)=ax^2+bx+c \\ \\ f(x)=(ax^2+bx)+c \\ \text{ look at these terms in the ( )} \\ \text{ we want to factor from it the coefficient of } x^2 \\ f(x)=(ax^2+\frac{a}{a} bx)+c \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ f(x)+a(?^2)=a(x^2+\frac{b}{a}x+?^2)+c \\ \text{ now I add } a(?^2) \text{ on both sides } \\ \text{ normally I just do everything on one side by adding a 0 } \\ \text{ this selection of the } ? \text{ will be so that we can complete the square for the ( ) part } \\ \text{ recall } x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \\ \text{ so we need } ?=\frac{b}{2a} \\ f(x)+a(\frac{b}{2a})^2=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c\] \[f(x)+a(\frac{b}{2a})^2=a(x+\frac{b}{2a})^2+c\] \[\text{ So normally I write in the form } y=a(x-h)^2+k\] \[f(x)=a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2 \\ \text{ where } h=\frac{-b}{2a} \\ \text{ and } k=c-a(\frac{b}{2a})^2 \\ \text{ where } (h,k) \text{ is the vertex }\]
Gabylovesyou
  • Gabylovesyou
y = 2(x - 1)^2 + 3 ?
freckles
  • freckles
did you cheat and use the formula above? :p
Gabylovesyou
  • Gabylovesyou
nope.. lol cx
freckles
  • freckles
\[y=2x^2-4x+5 \\ y=(2x^2-4x)+5 \\ y=2(x^2-2x)+5 \\ y+2(1)=2(x^2-2x+1)+5 \\\ y+2=2(x-1)^2+5 \\ y=2(x-1)^2+5-2 \\ y=2(x-1)^2+3 \] you could have used the formula above to check to see if your work provided you with the right answer though... \[y=ax^2+bx+c \\ y=a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2 \\ a=2,b=-4,c=5 \\ y=2(x+\frac{-4}{2(2)})^2+5-2(\frac{-4}{2(2)})^2 \\ y=2(x-1)^2+5-2(-1)^2 \\ y=2(x-1)^2+5-2 \\ y=2(x-1)^2+3\]

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