Complete the square and find the minimum value of the quadratic function 𝑦=3𝑥^2+12𝑥+20. (Give your answer as a whole or exact number.) 𝑦min=

Answer 1

y = 3x^2 + 12x + 20 = 0

x^2 + 4x = -20/3

x^2 +4x + 4 = 4-20/3 = 12/3 - 20/3 = - 8/3

(x +2)^2 = -8/3

take square roots of both sides

x+2 = + or - 2isqr(2/3)

x = -2 + 2isqr(2/3) or -2-2isqr(2/3)

x = -2 + 2isqr(2/3)

just checking the discriminant initially tells you there is no real solution, just imaginary ones

imaginary solutions mean the parabola never intersects the x axis, and the vertex is above the x axis with y coordinate > 0

any quadratic always has 2 solutions, or 1 with multiplicity 2

x =-2 is the axis of symmetry and the x coordinate of the vertex.

plug -2 into y = 3x^2 + 12x + 20 to get the y coordinate of the vertex

y = 3(4) -24+20 = 8

vertex and minimum point is (-2, 8)

or

just rewrite in vertex form y=a(x-h)^2 + k where (h,k) = the vertex

y = 3x^2 + 12x + 20

= 3(x^2 +4x +4) +20-12

= 3(x+2)^2 +8

it's an upward opening parabola with minimum point (-2,8) = vertex

Answer 2

Factor out a 3 from the first 2 terms.

y=3(x²+4x+c)+20

Now, determine the c value (b/2)²= (4/2)² = 4

Adding 4 in the parenthesis means adding 3x4=12 to the right side of the equation. So you have to also subtract 12 to keep the equation in balance.

y=3(x²+4x+4)+20-12

Simplify

y=3(x+2)²+8

The vertex, or the minimum is (-2,8)