Question

1. slide red circle in front of correct answer. determine which function has a range \\( \\{ y | y \leq 3 \\} \\). a \\( f ( x ) = - 2 x ^ { 3 } + 3 x ^ { 2 } + x - 1 \\) b. \\( f ( x ) = - 2 | x + 3 | - 3 \\) c \\( f ( x ) = - 2 ( x - 4 ) ^ { 2 } + 3 \\) d \\( f ( x ) = - 2 ^ { x - 1 } \\)

Explanation:

Step1: Analyze function \(a\)

Function \(a\): \(f(x)=-2x^{3}+3x^{2}+x - 1\) is a cubic function. The range of a cubic function \(y = ax^{3}+bx^{2}+cx + d\) (\(a
eq0\)) is \((-\infty,\infty)\) since \(\lim_{x
ightarrow\infty}(-2x^{3}+3x^{2}+x - 1)=-\infty\) and \(\lim_{x
ightarrow-\infty}(-2x^{3}+3x^{2}+x - 1)=\infty\).

Step2: Analyze function \(b\)

For the absolute - value function \(y = f(x)=-2|x + 3|-3\). The absolute - value part \(|x + 3|\geq0\). Multiply by \(- 2\): \(-2|x + 3|\leq0\). Then add \(-3\): \(y=-2|x + 3|-3\leq-3\). So the range is \(\{y|y\leq - 3\}\).

Step3: Analyze function \(c\)

For the quadratic function in vertex form \(y = f(x)=-2(x - 4)^{2}+3\). The square part \((x - 4)^{2}\geq0\). Multiply by \(-2\): \(-2(x - 4)^{2}\leq0\). Then add \(3\): \(y=-2(x - 4)^{2}+3\leq3\).

Step4: Analyze function \(d\)

For the exponential function \(y = f(x)=-2^{x - 1}\). The exponential function \(y = 2^{x-1}\) has a range of \((0,\infty)\). Multiply by \(-1\): \(y=-2^{x - 1}\) has a range of \((-\infty,0)\).

Answer:

C. \(f(x)=-2(x - 4)^{2}+3\)