what is the range of the function ( f(x)=3x^{2}+6x - 8 )?\n( { y|ygeq - 1} )\n( { y|yleq - 1} )\n( { y|ygeq…

what is the range of the function ( f(x)=3x^{2}+6x - 8 )?\n( { y|ygeq - 1} )\n( { y|yleq - 1} )\n( { y|ygeq - 11} )\n( { y|yleq - 11} )

what is the range of the function ( f(x)=3x^{2}+6x - 8 )?\n( { y|ygeq - 1} )\n( { y|yleq - 1} )\n( { y|ygeq - 11} )\n( { y|yleq - 11} )

Answer

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function (y = ax^{2}+bx + c), the (x) - coordinate of the vertex is (x=-\frac{b}{2a}). Given (f(x)=3x^{2}+6x - 8), where (a = 3), (b = 6), (c=-8). (x=-\frac{6}{2\times3}=-1)

Step2: Find the (y) - coordinate of the vertex

Substitute (x = - 1) into the function (f(x)): (f(-1)=3\times(-1)^{2}+6\times(-1)-8) (=3\times1-6 - 8) (=3-6 - 8=-11) Since (a = 3>0), the parabola opens upward.

Answer:

({y|y\geq - 11})