A quadratic function in standard form has a y-intercept of 12 and a non-real root of x=-2+2isqrt3. Write the function in both standard and vertex form.

Answer 1

Complex roots come in conjugate pairs.

Leave it to you to multiply out.

f(x) = x² + 4x + 16

y-intercept at x = 0. For the function the intercept is 16.

It cannot be made into 12.

Check your post for accuracy.

Answer 2

x= -2 + 2isqr3

x+2 + 2isqr3

square both sides to get

x^2+4x +4 = -12

x^2+4x +16 = 0 in ax^2+bx+c form, with a=1,b=4, c=16

or

(x+2)^2 +12 =0 in vertex form with vertex= (-2,12) = minimum of an upward opening parabola

x= -b/2a +/- (1/2a)sqr(b^2-4ac)

= -4/2 +/- (1/2)sqr(16-64)

=-1/2 +/-2sqr3 = the two zeros

y=12 when x=-2, y intercept is when x=0 which is y=16

to keep the same 2 (imaginary) x intercepts (zeros or solutions), and decrease the y intercept, flatten the parabola, by decreasing the coefficient of the x^2 term

reduce the coefficient to zero

(0)(x+2)^2 +12

then the y-intercept =12, but the "quadratic" has "degenerated" into a linear function, f(x) = 12

in standard form (0)(x^2+4x+4)+12 =f(x) = (0)x^2 + (0)x + 12

general standard form ax^2 +bx+c where a=b=0 and c=12