A quadratic function in standard form has a y-int. of 12 and a non-real root of x=-2+2i/sqrt3. Write function in both standard and vertex form

Answer 1

Despite the fact that this problem may have been entered incorrectly (which is probable), both Wyzant respondents, Mark M. and Raymond B. had partially incorrect conclusions in their solutions anyway.

Mark M. ignored the division symbol prior to the sqrt(3). Assuming that this is not incorrect, the solution would yield the equations: 3x^2+12x+16 and 3(x+2)^2+4 in general form and vertex form respectively. On the other hand, Raymond B. did arrive at these equations proving the point. Thus, the y-intercept is 16, not 12 as stated in the original problem. This ironically proves Mark M.'s final comment and question are appropriate.

But Raymond B. also mistakenly assumed that the roots, although still complex, would NOT change by scaling the “a” coefficient from 3 to 2. This is not the case as the altered roots would be:  -2+sqrt(2)i and -2-sqrt(2)i.

Answer 2

f(x) = ((x-(-2+2i√3))(x-(-2-2i√3))

f(x) = x² + 4x + 16, standard form

f(x) = (x² + 4x + 4) + 16 - 4

f(x) = (x + 2)² + 12, vertex form

Vertex: (-2, 12)

Not possible for the y-intercept to be at (0, 12).

I repeat, review the post for accuracy.

Answer 3

y intercept = 12

quadratic

zeros = -2 +/-2i/sqr3 (they come in conjugate pairs)

x = -2-2i/sqr3

x+2 = -2i/sqr3

square both sides

(x+2)^2 = -4/3

3(x+2)^2 = -4

3(x+2)^2 + 4 = 0

that has vertex (-2,4)

y intercept is when x=0

=3(0+2)^2 +4

= 3(2^2)+4

=3(4)+4

=12+4

= 16

that's close but not quite, if you require y intercept =12

flatten the parabola a little, make the coefficient of x^2 term, 3, a little smaller

try a =2= the x^2 term's coefficient, a(x-h)^2+k is the general form of the quadratic with (h,k)=vertex

2(x+2)^2 +4 = 0

for x=0

2(2^2) +4 =

2(4)+4

= 8+4

= 12

bingo

in vertex form

2(x+2)^2 + 4 where vertex = (-2,4)

in standard form multiply it out

2(x^2 + 4x +4) +4

= 2x^2 +8x +8 + 4

= 2x^2 + 8x + 12

it's an upward opening parabola with vertex=minimum point = (-2,4)

no x intercepts (that's why the zeros are imaginary) and y intercept when x=0 (0,12)

but that changes the zero from the given -2-2isqr3 to -2-2isqr2

close but not quite

graphing the parabola usually helps see what's going on, but it doesn't help see where the imaginary zero is

so, if you want the same zero, it is impossible, if you take a close zero, see above