Question

given the function ( g(t) = 3t^{4}-24t^{3}+45t^{2} ) its ( g )-intercept is enter a point more.. its ( t )-intercepts are

Explanation:

Step1: Find the \(g\)-intercept

The \(g\)-intercept occurs when \(t = 0\).
Substitute \(t=0\) into \(g(t)\):
\(g(0)=3\times0^{4}-24\times0^{3}+45\times0^{2}=0\)
So the \(g\)-intercept is the point \((0,0)\)

Step2: Find the \(t\)-intercepts

The \(t\)-intercepts occur when \(g(t) = 0\).
Set \(g(t)=3t^{4}-24t^{3}+45t^{2}=0\)
Factor out \(3t^{2}\): \(3t^{2}(t^{2}-8t + 15)=0\)
Factor the quadratic \(t^{2}-8t + 15=(t - 3)(t - 5)\)
So \(3t^{2}(t - 3)(t - 5)=0\)
Using the zero - product property:
\(3t^{2}=0\) gives \(t = 0\); \(t-3=0\) gives \(t=3\); \(t - 5=0\) gives \(t=5\)
The \(t\)-intercepts are the points \((0,0)\), \((3,0)\), \((5,0)\)

Answer:

The \(g\)-intercept is \((0,0)\). The \(t\)-intercepts are \((0,0)\), \((3,0)\), \((5,0)\)