Question$$f(x)=-3 x^{2}+2 x+3$$
Round to the nearest hundredth if necessary. If there is more than one $x$-intercept, separate them If applicable, click on "None". \begin{tabular}{|ll|} \hline vertex: & (II, $\square$ \\ $x$-intercept(s): & $\square$ \\ \hline \end{tabular}

Studdy Solution
Substitute the discriminant into the quadratic formula:
$$x = \frac{-2 \pm \sqrt{40}}{2(-3)}$$
$$= \frac{-2 \pm \sqrt{40}}{-6}$$
Simplify $\sqrt{40}$:
$$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$$
$$x = \frac{-2 \pm 2\sqrt{10}}{-6}$$
$$= \frac{-2}{-6} \pm \frac{2\sqrt{10}}{-6}$$
$$= \frac{1}{3} \mp \frac{\sqrt{10}}{3}$$
The $x$-intercepts are:
$$x = \frac{1 + \sqrt{10}}{3} \quad \text{and} \quad x = \frac{1 - \sqrt{10}}{3}$$
Calculate these values to the nearest hundredth:
$$x \approx \frac{1 + 3.16}{3} \approx \frac{4.16}{3} \approx 1.39$$
$$x \approx \frac{1 - 3.16}{3} \approx \frac{-2.16}{3} \approx -0.72$$
The vertex is $\left(\frac{1}{3}, \frac{10}{3}\right) \approx (0.33, 3.33)$.
The $x$-intercepts are approximately $1.39$ and $-0.72$.