Express the quadratic functions in standard form:1)x^2-4x
2)2x^2+6x
3)5x^2+10x+6
4)2x^2-16+39
5)x^2+8x
6)2x^2+8x+14
7)3-12x-12x^2.Well, the standard form of a quadratic function is
f(x)=a(x-h)^2+k
where h and k are numbers.The standard procedure for writing a function in standard form is called completnig the square.
It goes like this:You work with x^2 – bx
You compare it to the first two terms in the expansion (x-h)^2 = x^2 -2hx +h^2
Then, you conclude that h must be b/2 to have the coefficients of x in (x^2 – bx) and (x^2 – 2hx +h^2) equal.If h = b/2, then expanding (x – b/2)^2 = x^2 – bx +(b/2)^2, we notice that our expression, x^2 – bx is (x-b/2)^2 – (b/2)^2
The compensatory term (b/2)^2 can also be written as b^2/4And so, we have the formula:x^2 – bx = (x – b/2)^2 – b^2/4Now, lets see:1) x^2-4x = (x – 2)^2 – 4, becasue b=4 here, b/2 = 2, and 2^2 = 4
2) factor our 2 first: 2(x^2+6x), then write 2(x^2 + 3x) as 2(x^2 – (-3)x). We have: 2x^2 + 6x = 2(x-[-3/2])^2 – [-3/2]^2 = 2(x+1.5)^2 – 9/4
3) factor out 5 from the first two terms: 5(x^2+2x)+6 = 5([x+1]^2 – 1)+6 Open parentheses and combine numbers:
5(x+1)^2 – 5 + 6 = 5(x+1)^2+1
4) Im assuming you meant 2x^2-16x+39; factor out 2 from the first two terms: 2(x^2-8x) + 39. Complete the square inside parentheses: 2([x-4]^2-4^2)+39 = 2(x-4)^2-32+39 = 2(x-4)^2+7
5) (x+4)^2 – 64
6) 2(x^2+4x)+14 = 2([x+2]^2-4)+14 = 2(x+2)^2-8+14 = 2(x+2)^2+6
7) factor out (-12) and write the expression as (-12)(x^2+1)+3=(-12)([x+1/2]^2-1/4)+3. Open parentheses: -12(x+1/2)^2 + 12/4+3
=-12(x+1/2)^2+6.
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