tell whether the function has a minimum value or a maximum value. then find the value.\n9. ( y = 3 x ^ { 2 }…

tell whether the function has a minimum value or a maximum value. then find the value.\n9. ( y = 3 x ^ { 2 } - 12 x + 1 )\n10. ( y = - 4 x ^ { 2 } + 48 x - 144 )\n11. ( y = - \frac { 1 } { 2 } x ^ { 2 } - 8 x - 7 )\n12. ( y = 2 x ^ { 2 } + 2 x + 7 )

tell whether the function has a minimum value or a maximum value. then find the value.\n9. ( y = 3 x ^ { 2 } - 12 x + 1 )\n10. ( y = - 4 x ^ { 2 } + 48 x - 144 )\n11. ( y = - \frac { 1 } { 2 } x ^ { 2 } - 8 x - 7 )\n12. ( y = 2 x ^ { 2 } + 2 x + 7 )

Answer

Explanation:

Step1: Determine the form of the function

For a quadratic function (y = ax^{2}+bx + c), if (a>0), the function has a minimum value; if (a < 0), the function has a maximum value.

Step2: Analyze function (y = 3x^{2}-12x + 1)

Here (a = 3), (b=-12), (c = 1). Since (a=3>0), the function has a minimum value. The (x) - coordinate of the vertex (where the minimum occurs) is given by (x=-\frac{b}{2a}). Substitute (a = 3) and (b=-12) into (x =-\frac{b}{2a}): (x=-\frac{-12}{2\times3}=\frac{12}{6} = 2) Substitute (x = 2) into (y=3x^{2}-12x + 1): (y=3\times(2)^{2}-12\times2 + 1=3\times4-24 + 1=12-24 + 1=-11)

Step3: Analyze function (y=-4x^{2}+48x-144)

Here (a=-4), (b = 48), (c=-144). Since (a=-4<0), the function has a maximum value. The (x) - coordinate of the vertex (where the maximum occurs) is (x=-\frac{b}{2a}). Substitute (a=-4) and (b = 48) into (x=-\frac{b}{2a}): (x=-\frac{48}{2\times(-4)}=\frac{-48}{-8}=6) Substitute (x = 6) into (y=-4x^{2}+48x-144): (y=-4\times(6)^{2}+48\times6-144=-4\times36 + 288-144=-144+288-144 = 0)

Step4: Analyze function (y=-\frac{1}{2}x^{2}-8x-7)

Here (a=-\frac{1}{2}), (b=-8), (c=-7). Since (a=-\frac{1}{2}<0), the function has a maximum value. The (x) - coordinate of the vertex is (x=-\frac{b}{2a}). Substitute (a=-\frac{1}{2}) and (b=-8) into (x=-\frac{b}{2a}): (x=-\frac{-8}{2\times(-\frac{1}{2})}=-\frac{-8}{-1}=-8) Substitute (x=-8) into (y=-\frac{1}{2}x^{2}-8x-7): (y=-\frac{1}{2}\times(-8)^{2}-8\times(-8)-7=-\frac{1}{2}\times64 + 64-7=-32+64-7=25)

Step5: Analyze function (y=2x^{2}+2x + 7)

Here (a = 2), (b=2), (c = 7). Since (a=2>0), the function has a minimum value. The (x) - coordinate of the vertex is (x=-\frac{b}{2a}). Substitute (a = 2) and (b=2) into (x=-\frac{b}{2a}): (x=-\frac{2}{2\times2}=-\frac{1}{2}) Substitute (x=-\frac{1}{2}) into (y=2x^{2}+2x + 7): (y=2\times(-\frac{1}{2})^{2}+2\times(-\frac{1}{2})+7=2\times\frac{1}{4}-1 + 7=\frac{1}{2}-1+7=\frac{1 - 2+14}{2}=\frac{13}{2}=6.5)

Answer:

  • For (y = 3x^{2}-12x + 1): The function has a minimum value of (-11).
  • For (y=-4x^{2}+48x-144): The function has a maximum value of (0).
  • For (y=-\frac{1}{2}x^{2}-8x-7): The function has a maximum value of (25).
  • For (y=2x^{2}+2x + 7): The function has a minimum value of (6.5).