How do I write these into standard quadratic form? y= 3(x-2)^2-5 f(x)= -2(x+4)^2+3
it'd be good to look up "standard quadratic form." I'd suggest you do that. I'll drop you a hint: See http://www.mathsisfun.com/algebra/quadratic-equation.html In the first case, \[y=3(x-2)^2 - 5\] should be re-written by doing the following: 1. expand (x-2)^2. 2. Multiply each resulting term by 3. 3. Subtract 5 from the result. 4. combine like terms and re-arrange them in descending order by power of x.
\[3x ^{2}-17\] ?!
Would you mind showing how you obtain your answer, please? I'd be especially interested in seeing how you expand (x-2)^2.
\[y=3(x-2)^{2}-5\] \[(x-2)^{2}=x ^{2}-4\] \[3(x ^{2}-4)=3x ^{2}-12\] \[3x ^{2}-12-5=3x ^{2}-17\]
(x-2)^2 = x^2 -4x + 4
How do you get that? @douglaswinslowcooper
\[(x-2)^{2}=x ^{2}-4\]is incorrect. Please recall that (x-2)^2 signifies (x-2)(x-2). Try again, please.
Important: Please show your work here.
Do I do distributive property?
That's certainly part of what you need to do. Many students refer to the "FOIL" method of multiplying binomials together.
What's FOIL? I'm used to PEMDAS. (Parentheses, Exponents, Multiplying, Division, Addition, Subtraction)
(x-2)(x-2) = x(x-2) - 2(x-2) = x^2-2x-2x+4=x^2-4x+4
I don't get that.
FOIL refers to First, Outer, Inner, Last. In (x-2)(x-2), the First terms are x and x. Multiplying them together, we get x^2. The Outer terms are x and -2. The Inner terms are -2 and x. The Last terms are -2 and -2. See this before?
No.
douglaswinslowcooper has a different method to multiply (x-2)(x-2), but his result is correct. In summary, you very much do need to know how to multiply binomials. Ask for further clarification if you need it.
Do you have an algebra textbook?
Yes.
Look up "special products and factoring." Or look at http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut29_specfact.htm.
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