Step-by-step explanation:
(a-b)^2=(a^2-2*a*b+b^2)
4*(x^2-2x+1)-9=4x^2-8x+4-9=4x^2-8x-5
if you have to calculate the Δ
there is it: 4x^2-8x-5=0
Δ=b^2-4*a*c=64-4*4*(-5)=64+80=144
-b±Δ/2*a=8±12/8=
=8+12/8=20/8|:4=5/2
=8-12/8=-4/8|:4=-1/2
Answer:
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Step-by-step explanation:
Answer:
6n
Step-by-step explanation:
If I'm reading your equations correctly, they are:f(x)=x2-8x+15g(x)=x-3h(x)=f(x)/g(x)The domain of a function is the set of all possible inputs, what we can plug in for our variable.The largest two limitations on domains (other than explicit limitations, like in piecewise functions) are radicals and rational functions. With radical expressions we know that we CANNOT take an even root of a negative number. I don't see that problem here. With rationals we know that we CANNOT divide by zero. So the question becomes, when does h(x) ask us to divide by zero? When is the denominator of h(x) zero?Since the denominator of h(x) is g(x), we cannot let g(x) equal zero. So when does that happen? when x-3=0 or when x=3. I hope you see here that if x=3, then g(x)=0, and so h(x)=f(x)/0, which we CANNOT do. The domain of h(x) is all real numbers not equal to 3. There is more going on here. If you had factored f(x) first, you could have written h(x) in a confusing way:h(x)=( f(x) ) / ( g(x) )h(x)= ( (x-5)(x-3) ) / (x-3) Right here, it looks like (x-3) will cancel out from the top and bottom of your fraction. It does, in a way. The graph of h(x) will behave exactly like the line y=x-5, except that it has a hole in it at x=3 (check this! it's cool!)SOOO, the takeaway is that it is better to determine limitations on your domain BEFORE over-simplifying your equations.
