Determine the vertical asymptotes of the graph of the function. Write numbers as simplified fractions or integers. Give your answers in exact form. $$ g(t)=\frac{t-1}{2 t^{2}+3 t-2} $$ Part: $$ 0 / 3 $$ Part 1 of 3 Factor the numerator and denominator. $$ g(t)=\frac{t-1}{\text { [1] }} $$

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It seems that the expression $$t-1$$ cannot be factored further. Let's move on to the next step.
Factor the expression by following steps:
- step0: Factor:
  $$2t^{2}+3t-2$$
- step1: Rewrite the expression:
  $$2t^{2}+\left(-1+4\right)t-2$$
- step2: Calculate:
  $$2t^{2}-t+4t-2$$
- step3: Rewrite the expression:
  $$t\times 2t-t+2\times 2t-2$$
- step4: Factor the expression:
  $$t\left(2t-1\right)+2\left(2t-1\right)$$
- step5: Factor the expression:
  $$\left(t+2\right)\left(2t-1\right)$$
The denominator $$2t^2 + 3t - 2$$ can be factored as $$(t+2)(2t-1)$$.

Now, let's rewrite the function $$g(t)$$ with the factored numerator and denominator:

$$ g(t) = \frac{t-1}{(t+2)(2t-1)} $$

To find the vertical asymptotes, we need to find the values of $$t$$ that make the denominator equal to zero. The denominator will be zero when either $$t+2=0$$ or $$2t-1=0$$.

Solving for $$t$$ in both cases:

1. $$t+2=0$$ gives $$t=-2$$.
2. $$2t-1=0$$ gives $$t=\frac{1}{2}$$.

Therefore, the vertical asymptotes of the graph of the function $$g(t)$$ are at $$t=-2$$ and $$t=\frac{1}{2}$$.
The vertical asymptotes of the function $$ g(t) = \frac{t-1}{2t^2 + 3t - 2} $$ are at $$ t = -2 $$ and $$ t = \frac{1}{2} $$.

Determine the vertical asymptotes of the graph of the function. Write numbers as simplified fractions or integers. Give your answers in exact form. \[ g(t)=\frac{t-1}{2 t^{2}+3 t-2} \] Part: \( 0 / 3 \) Part 1 of 3 Factor the numerator and denominator. \[ g(t)=\frac{t-1}{\text { [1] }} \]

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